{"id":4223,"date":"2023-02-02T06:01:00","date_gmt":"2023-02-02T06:01:00","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4223"},"modified":"2023-02-02T06:03:39","modified_gmt":"2023-02-02T06:03:39","slug":"ncert-solutions-class-9-maths-chapter-2-polynomials-ex-2-5","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-2-polynomials-ex-2-5\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.5"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 9 Maths\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 2 (Polynomials)\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 9 Mathematics\u00a0<strong>Chapter &#8211; 2 Polynomials <\/strong>Exercise 2.5 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Chapter 2: Polynomials\u00a0<\/strong><\/span><\/p>\n<ul>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-2-polynomials-ex-2-1\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 2.1<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-2-polynomials-ex-2-2\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 2.2<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-2-polynomials-ex-2-3\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 2.3<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-2-polynomials-ex-2-4\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 2.4<\/a><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 2.5\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. Use suitable identities to find the following products:<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong>(i) (x + 4)(x + 10)<br \/>\n(ii) (x + 8) (x &#8211; 10)<br \/>\n(iii) (3x + 4) (3x \u2013 5)<br \/>\n(iv) (y<sup>2 <\/sup>+ <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{3}{2}}\" alt=\"\\mathbf{\\frac{3}{2}}\" align=\"absmiddle\" \/>) (y<sup>2 <\/sup>\u2013 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{3}{2}}\" alt=\"\\mathbf{\\frac{3}{2}}\" align=\"absmiddle\" \/>)<br \/>\n(v) (3 \u2013 2x) (3 + 2x)<br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><strong><span style=\"color: #000000;\">Answer &#8211; <\/span><\/strong><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) (x + 4)(x + 10)<br \/>\n<\/strong>Using the identity, (x + a)(x + b) = x <sup>2<\/sup>+(a + b)x + ab<br \/>\n<\/span><span style=\"color: #000000;\">Here, a = 4 and b = 10<br \/>\n<\/span><span style=\"color: #000000;\">We get,<br \/>\n<\/span><span style=\"color: #000000;\">(x + 4)(x + 10)<br \/>\n= x<sup>2 <\/sup>+ (4 + 10)x + (4\u00d710)<br \/>\n<\/span><span style=\"color: #000000;\">= x<sup>2 <\/sup>+ 14x + 40<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) (x + 8)(x \u2013 10)<br \/>\n<\/strong><\/span><span style=\"color: #000000;\">Using the identity, (x + a)(x + b) = x <sup>2 <\/sup>+ (a + b)x + ab<br \/>\n<\/span><span style=\"color: #000000;\">Here, a = 8 and b = \u221210<br \/>\n<\/span><span style=\"color: #000000;\">We get,<br \/>\n<\/span><span style=\"color: #000000;\">(x + 8)(x \u2212 10)<br \/>\n= x<sup>2 <\/sup>+ (8 + (\u221210))x + (8\u00d7(\u221210))<br \/>\n<\/span><span style=\"color: #000000;\">= x<sup>2 <\/sup>+ (8 \u2212 10)x \u2013 80<br \/>\n<\/span><span style=\"color: #000000;\">= x<sup>2 <\/sup>\u2212 2x \u2212 80<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) (3x + 4)(3x \u2013 5)<br \/>\n<\/strong><\/span><span style=\"color: #000000;\">Using the identity, (x + a)(x + b) = x<sup>2 <\/sup>+ (a + b)x + ab<br \/>\n<\/span><span style=\"color: #000000;\">Here, x = 3x, a = 4 and b = \u22125<br \/>\n<\/span><span style=\"color: #000000;\">We get,<br \/>\n<\/span><span style=\"color: #000000;\">(3x + 4)(3x \u2212 5)<br \/>\n= (3x)<sup>2 <\/sup>+ [4+(\u22125)]3x + 4\u00d7(\u22125)<br \/>\n<\/span><span style=\"color: #000000;\">= 9x<sup>2<\/sup>+ 3x(4\u20135) \u2013 20<br \/>\n<\/span><span style=\"color: #000000;\">= 9x<sup>2 <\/sup>\u2013 3x \u2013 20<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv) (y<sup>2 <\/sup>+ <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{3}{2}}\" alt=\"\\mathbf{\\frac{3}{2}}\" align=\"absmiddle\" \/>)(y<sup>2 <\/sup>&#8211; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{3}{2}}\" alt=\"\\mathbf{\\frac{3}{2}}\" align=\"absmiddle\" \/>)<br \/>\n<\/strong><\/span><span style=\"color: #000000;\">Using the identity, (x + y)(x \u2013 y) = x<sup>2 <\/sup>\u2013 y<sup>2<br \/>\n<\/sup><\/span><span style=\"color: #000000;\">Here, x = y<sup>2<\/sup>and y = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{3}{2}\" alt=\"\\frac{3}{2}\" align=\"absmiddle\" \/><br \/>\n<\/span><span style=\"color: #000000;\">We get,<br \/>\n<\/span><span style=\"color: #000000;\">(y<sup>2 <\/sup>+ <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{3}{2}\" alt=\"\\frac{3}{2}\" align=\"absmiddle\" \/>)(y<sup>2 <\/sup>\u2013 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{3}{2}\" alt=\"\\frac{3}{2}\" align=\"absmiddle\" \/>)\u00a0 <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">= (y<sup>2<\/sup>)<sup>2 <\/sup>\u2013 (<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{3}{2}\" alt=\"\\frac{3}{2}\" align=\"absmiddle\" \/>)<sup>2 <\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">= y<sup>4 <\/sup>\u2013 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{9}{4}\" alt=\"\\frac{9}{4}\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. Evaluate the following products without multiplying directly:<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong>(i) 103 \u00d7 107<br \/>\n(ii) 95 \u00d7 96<br \/>\n(iii) 104 \u00d7 96<br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<\/strong><br \/>\n<\/span><span style=\"color: #000000;\"><strong>(i) 103 \u00d7 107<br \/>\n<\/strong>103 \u00d7 107 = (100 + 3) \u00d7 (100 + 7)<br \/>\n<\/span><span style=\"color: #000000;\">Using identity, [(x + a)(x + b) = x<sup>2 <\/sup>+ (a + b)x + ab]<br \/>\n<\/span><span style=\"color: #000000;\">Here, x = 100, <\/span><span style=\"color: #000000;\">a = 3 and\u00a0<\/span><span style=\"color: #000000;\">b = 7<br \/>\n<\/span><span style=\"color: #000000;\">We get,<br \/>\n103 \u00d7 107 = (100+3)\u00d7(100+7)<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2 (100)<sup>2 <\/sup>+ (3+7)100+(3\u00d77)<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2 10000 + 1000 + 21<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2 11021<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) 95 \u00d7 96<br \/>\n<\/strong><\/span><span style=\"color: #000000;\">95 \u00d7 96 = (100 &#8211; 5)\u00d7(100 &#8211; 4)<br \/>\n<\/span><span style=\"color: #000000;\">Using identity, [(x &#8211; a)(x &#8211; b) = x<sup>2 <\/sup>&#8211; (a + b)x + ab]<br \/>\n<\/span><span style=\"color: #000000;\">Here, x = 100, <\/span><span style=\"color: #000000;\">a = -5 and\u00a0<\/span><span style=\"color: #000000;\">b = -4<br \/>\n<\/span><span style=\"color: #000000;\">We get,<br \/>\n95 \u00d7 96 = (100 &#8211; 5)\u00d7(100 &#8211; 4)<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2 (100)<sup>2 <\/sup>+ 100(-5 + (-4)) + (-5\u00d7-4)<br \/>\n<\/span><span style=\"color: #000000;\">= 10000 &#8211; 900 + 20<br \/>\n<\/span><span style=\"color: #000000;\">= 9120<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) 104 \u00d7 96<br \/>\n<\/strong><\/span><span style=\"color: #000000;\">104 \u00d7 96 = (100 + 4)\u00d7(100 \u2013 4)<br \/>\n<\/span><span style=\"color: #000000;\">Using identity, [(a + b)(a &#8211; b)= a<sup>2 <\/sup>&#8211; b<sup>2<\/sup>]<br \/>\n<\/span><span style=\"color: #000000;\">Here, a = 100, <\/span><span style=\"color: #000000;\">b = 4<br \/>\n<\/span><span style=\"color: #000000;\">We get,<br \/>\n104 \u00d7 96 = (100 + 4)\u00d7(100 \u2013 4)<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2 (100)<sup>2 <\/sup>\u2013 (4)<sup>2<br \/>\n<\/sup><\/span><span style=\"color: #000000;\">\u21d2 10000 \u2013 16<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2 9984<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. Factorize the following using appropriate identities:<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong>(i) 9x<sup>2 <\/sup>+ 6xy + y<sup>2<br \/>\n<\/sup>(ii) 4y<sup>2 <\/sup>&#8211; 4y + 1<br \/>\n(iii) x<sup>2<\/sup> \u2013 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{y^2}{100}}\" alt=\"\\mathbf{\\frac{y^2}{100}}\" align=\"absmiddle\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<\/strong><br \/>\n<\/span><span style=\"color: #000000;\"><strong>(i) 9x<sup>2 <\/sup>+ 6xy + y<sup>2<br \/>\n<\/sup><\/strong><\/span><span style=\"color: #000000;\">= (3x)<sup>2 <\/sup>+ (2 \u00d7 3x \u00d7 y) + y<sup>2<br \/>\n<\/sup><\/span><span style=\"color: #000000;\">Using identity, x<sup>2 <\/sup>+ 2xy + y<sup>2\u00a0<\/sup>= (x + y)<sup>2<br \/>\n<\/sup><\/span><span style=\"color: #000000;\">Here, x = 3x,\u00a0<\/span><span style=\"color: #000000;\">y = y<br \/>\n<\/span><span style=\"color: #000000;\">9x<sup>2 <\/sup>+ 6xy + y<sup>2<br \/>\n<\/sup>= (3x)<sup>2 <\/sup>+ (2 \u00d7 3x \u00d7 y) + y<sup>2<br \/>\n<\/sup><\/span><span style=\"color: #000000;\">= (3x + y)<sup>2<br \/>\n<\/sup><\/span><span style=\"color: #000000;\">= (3x + y) (3x + y)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) 4y<sup>2<\/sup>\u22124y + 1<br \/>\n<\/strong><\/span><span style=\"color: #000000;\">= (2y)<sup>2<\/sup>\u2013(2 \u00d7 2y \u00d7 1) + 1<br \/>\n<\/span><span style=\"color: #000000;\">Using identity, x<sup>2<\/sup>\u00a0\u2013 2xy + y<sup>2\u00a0<\/sup>= (x \u2013 y)<sup>2<br \/>\n<\/sup><\/span><span style=\"color: #000000;\">Here, x = 2y,\u00a0<\/span><span style=\"color: #000000;\">y = 1<br \/>\n<\/span><span style=\"color: #000000;\">4y<sup>2<\/sup>\u22124y+1<br \/>\n= (2y)<sup>2 <\/sup>\u2013 (2 \u00d7 2y \u00d7 1) + 1<sup>2<br \/>\n<\/sup><\/span><span style=\"color: #000000;\">= (2y \u2013 1)<sup>2<br \/>\n<\/sup><\/span><span style=\"color: #000000;\">= (2y \u2013 1) (2y \u2013 1)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) x<sup>2<\/sup> \u2013 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{y^2}{100}}\" alt=\"\\mathbf{\\frac{y^2}{100}}\" align=\"absmiddle\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">= x<sup>2 <\/sup>\u2013 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(&amp;space;\\frac{y}{10}&amp;space;\\right&amp;space;)^2\" alt=\"\\left ( \\frac{y}{10} \\right )^2\" align=\"absmiddle\" \/><br \/>\n<\/span><span style=\"color: #000000;\">Using identity, x<sup>2 <\/sup>&#8211; y<sup>2\u00a0<\/sup>= (x &#8211; y)(x + y)<br \/>\n<\/span><span style=\"color: #000000;\">Here, x = x,\u00a0<\/span><span style=\"color: #000000;\">y = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{y}{10}\" alt=\"\\frac{y}{10}\" align=\"absmiddle\" \/><br \/>\n<\/span><span style=\"color: #000000;\">x<sup>2 <\/sup>\u2013 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{y^2}{100}\" alt=\"\\frac{y^2}{100}\" align=\"absmiddle\" \/>\u00a0 <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">= x<sup>2 <\/sup>\u2013 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(&amp;space;\\frac{y}{10}&amp;space;\\right&amp;space;)^2\" alt=\"\\left ( \\frac{y}{10} \\right )^2\" align=\"absmiddle\" \/> <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">= (x \u2013 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{y}{10}\" alt=\"\\frac{y}{10}\" align=\"absmiddle\" \/>) (x + <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{y}{10}\" alt=\"\\frac{y}{10}\" align=\"absmiddle\" \/>)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. Expand each of the following, using suitable identities:<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong>(i) (x + 2y + 4z)<sup>2<br \/>\n<\/sup><\/strong><\/span><span style=\"color: #000000;\"><strong>(ii) (2x \u2212 y + z)<sup>2<br \/>\n<\/sup><\/strong><\/span><span style=\"color: #000000;\"><strong>(iii) (\u22122x + 3y + 2z)<sup>2<br \/>\n<\/sup><\/strong><\/span><span style=\"color: #000000;\"><strong>(iv) (3a \u2013 7b \u2013 c)<sup>2<br \/>\n<\/sup><\/strong><\/span><span style=\"color: #000000;\"><strong>(v) (\u20132x + 5y \u2013 3z)<sup>2<br \/>\n<\/sup><\/strong><\/span><span style=\"color: #000000;\"><strong>(vi) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\left&amp;space;[&amp;space;\\frac{1}{4}a&amp;space;-&amp;space;\\frac{1}{2}b&amp;space;+1\\right&amp;space;]}^2\" alt=\"\\mathbf{\\left [ \\frac{1}{4}a - \\frac{1}{2}b +1\\right ]}^2\" align=\"absmiddle\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<\/strong><br \/>\n<\/span><span style=\"color: #000000;\"><strong>(i) (x + 2y + 4z)<sup>2<br \/>\n<\/sup><\/strong><\/span><span style=\"color: #000000;\">Using identity, (x + y + z)<sup>2<\/sup>\u00a0= x<sup>2 <\/sup>+ y<sup>2 <\/sup>+ z<sup>2 <\/sup>+ 2xy + 2yz + 2zx<br \/>\n<\/span><span style=\"color: #000000;\">Here, x = x,\u00a0<\/span><span style=\"color: #000000;\">y = 2y and\u00a0<\/span><span style=\"color: #000000;\">z = 4z<br \/>\n<\/span><span style=\"color: #000000;\">(x + 2y + 4z)<sup>2<br \/>\n<\/sup>= x<sup>2 <\/sup>+ (2y)<sup>2 <\/sup>+ (4z)<sup>2 <\/sup>+ (2 \u00d7 x \u00d7 2y) + (2 \u00d7 2y \u00d7 4z) + (2 \u00d7 4z \u00d7 x)<br \/>\n<\/span><span style=\"color: #000000;\">= x<sup>2 <\/sup>+ 4y<sup>2 <\/sup>+ 16z<sup>2 <\/sup>+ 4xy +16yz + 8xz<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) (2x \u2212 y + z)<sup>2<\/sup><br \/>\n<\/strong><\/span><span style=\"color: #000000;\">Using identity, (x + y + z)<sup>2<\/sup>\u00a0= x<sup>2 <\/sup>+ y<sup>2 <\/sup>+ z<sup>2 <\/sup>+ 2xy + 2yz + 2zx<br \/>\n<\/span><span style=\"color: #000000;\">Here, x = 2x, <\/span><span style=\"color: #000000;\">y = \u2212y,\u00a0<\/span><span style=\"color: #000000;\">z = z<br \/>\n<\/span><span style=\"color: #000000;\">(2x \u2212 y + z)<sup>2<br \/>\n<\/sup>= (2x)<sup>2 <\/sup>+ (\u2212y)<sup>2 <\/sup>+ z<sup>2 <\/sup>+ (2 \u00d7 2x \u00d7 \u2212y) + (2 \u00d7 \u2212y \u00d7 z) + (2 \u00d7 z \u00d7 2x)<br \/>\n<\/span><span style=\"color: #000000;\">= 4x<sup>2 <\/sup>+ y<sup>2 <\/sup>+ z<sup>2 <\/sup>\u2013 4xy \u2013 2yz + 4xz<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) (\u22122x + 3y + 2z)<sup>2<br \/>\n<\/sup><\/strong><\/span><span style=\"color: #000000;\">Using identity, (x + y + z)<sup>2<\/sup>\u00a0= x<sup>2 <\/sup>+ y<sup>2 <\/sup>+ z<sup>2 <\/sup>+ 2xy + 2yz + 2zx<br \/>\n<\/span><span style=\"color: #000000;\">Here, x = \u22122x,\u00a0<\/span><span style=\"color: #000000;\">y = 3y and\u00a0<\/span><span style=\"color: #000000;\">z = 2z<br \/>\n<\/span><span style=\"color: #000000;\">(\u22122x + 3y + 2z)<sup>2<br \/>\n<\/sup>= (\u22122x)<sup>2 <\/sup>+ (3y)<sup>2 <\/sup>+ (2z)<sup>2 <\/sup>+ (2 \u00d7\u22122x \u00d7 3y) + (2 \u00d7 3y \u00d7 2z) + (2 \u00d7 2z \u00d7 \u22122x)<br \/>\n<\/span><span style=\"color: #000000;\">= 4x<sup>2 <\/sup>+ 9y<sup>2 <\/sup>+ 4z<sup>2 <\/sup>\u2013 12xy + 12yz \u2013 8xz<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv) (3a \u2013 7b \u2013 c)<sup>2<br \/>\n<\/sup><\/strong><\/span><span style=\"color: #000000;\">Using identity (x + y + z)<sup>2<\/sup>\u00a0= x<sup>2 <\/sup>+ y<sup>2 <\/sup>+ z<sup>2 <\/sup>+ 2xy + 2yz + 2zx<br \/>\n<\/span><span style=\"color: #000000;\">Here, x = 3a, <\/span><span style=\"color: #000000;\">y = \u2013 7b and\u00a0<\/span><span style=\"color: #000000;\">z = \u2013 c<br \/>\n<\/span><span style=\"color: #000000;\">(3a \u2013 7b \u2013 c)<sup>2<br \/>\n<\/sup>= (3a)<sup>2 <\/sup>+ (\u2013 7b)<sup>2 <\/sup>+ (\u2013 c)<sup>2 <\/sup>+ (2 \u00d7 3a \u00d7\u2013 7b) + (2\u00d7\u2013 7b \u00d7\u2013 c) + (2\u00d7 \u2013c \u00d7 3a)<br \/>\n<\/span><span style=\"color: #000000;\">= 9a<sup>2<\/sup>\u00a0+ 49b<sup>2\u00a0<\/sup>+ c<sup>2<\/sup>\u2013 42ab + 14bc \u2013 6ca<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(v) (\u20132x+5y\u20133z)<sup>2<br \/>\n<\/sup><\/strong><\/span><span style=\"color: #000000;\">Using identity, (x + y + z)<sup>2<\/sup>\u00a0= x<sup>2 <\/sup>+ y<sup>2 <\/sup>+ z<sup>2 <\/sup>+ 2xy + 2yz + 2zx<br \/>\n<\/span><span style=\"color: #000000;\">Here, x = \u20132x, <\/span><span style=\"color: #000000;\">y = 5y and\u00a0<\/span><span style=\"color: #000000;\">z = \u2013 3z<br \/>\n<\/span><span style=\"color: #000000;\">(\u20132x + 5y \u2013 3z)<sup>2\u00a0<\/sup>= (\u20132x)<sup>2 <\/sup>+ (5y)<sup>2 <\/sup>+ (\u20133z)<sup>2 <\/sup>+ (2 \u00d7 \u20132x \u00d7 5y) + (2 \u00d7 5y \u00d7\u20133z) + (2 \u00d7 \u20133z \u00d7\u20132x)<br \/>\n<\/span><span style=\"color: #000000;\">= 4x<sup>2 <\/sup>+ 25y<sup>2\u00a0<\/sup>+ 9z<sup>2 <\/sup>\u2013 20xy \u2013 30yz + 12zx<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(vi) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\left&amp;space;[&amp;space;\\frac{1}{4}a&amp;space;-&amp;space;\\frac{1}{2}b&amp;space;+1\\right&amp;space;]}^2\" alt=\"\\mathbf{\\left [ \\frac{1}{4}a - \\frac{1}{2}b +1\\right ]}^2\" align=\"absmiddle\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">Using identity, (x + y + z)<sup>2<\/sup>\u00a0= x<sup>2 <\/sup>+ y<sup>2 <\/sup>+ z<sup>2 <\/sup>+ 2xy + 2yz + 2zx<br \/>\n<\/span><span style=\"color: #000000;\">Here, x = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{4}\" alt=\"\\frac{1}{4}\" align=\"absmiddle\" \/>a, <\/span><span style=\"color: #000000;\">y = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?-\\frac{1}{2}\" alt=\"-\\frac{1}{2}\" align=\"absmiddle\" \/>b and <\/span><span style=\"color: #000000;\">z = 1<br \/>\n<strong><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;[&amp;space;\\frac{1}{4}a&amp;space;-&amp;space;\\frac{1}{2}b&amp;space;+1\\right&amp;space;]^2\" alt=\"\\left [ \\frac{1}{4}a - \\frac{1}{2}b +1\\right ]^2\" align=\"absmiddle\" \/> <\/strong><br \/>\n<\/span><\/p>\n<p>= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(&amp;space;\\frac{1}{4}a&amp;space;\\right&amp;space;)^2&amp;space;+&amp;space;\\left&amp;space;(&amp;space;\\frac{1}{2}b&amp;space;\\right&amp;space;)^2&amp;space;+&amp;space;1^2&amp;space;+&amp;space;\\left&amp;space;(&amp;space;2\\times&amp;space;\\frac{1}{4}a\\times&amp;space;-\\frac{1}{2}b\\right&amp;space;)&amp;space;+&amp;space;\\left&amp;space;(&amp;space;2\\times&amp;space;-\\frac{1}{2}b\\times&amp;space;1\\right)&amp;space;+&amp;space;\\left&amp;space;(&amp;space;2\\times&amp;space;1\\times&amp;space;\\frac{1}{4}a\\right&amp;space;)\" alt=\"\\left ( \\frac{1}{4}a \\right )^2 + \\left ( \\frac{1}{2}b \\right )^2 + 1^2 + \\left ( 2\\times \\frac{1}{4}a\\times -\\frac{1}{2}b\\right ) + \\left ( 2\\times -\\frac{1}{2}b\\times 1\\right) + \\left ( 2\\times 1\\times \\frac{1}{4}a\\right )\" align=\"absmiddle\" \/><\/p>\n<p>= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{16}a^2&amp;space;+&amp;space;\\frac{1}{4}b^2&amp;space;+&amp;space;1&amp;space;-&amp;space;\\frac{2}{8}ab&amp;space;-\\frac{2}{2}b&amp;space;+\\frac{2}{4}a\" alt=\"\\frac{1}{16}a^2 + \\frac{1}{4}b^2 + 1 - \\frac{2}{8}ab -\\frac{2}{2}b +\\frac{2}{4}a\" align=\"absmiddle\" \/><\/p>\n<p>= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{16}a^2&amp;space;+&amp;space;\\frac{1}{4}b^2&amp;space;+1&amp;space;-&amp;space;\\frac{1}{4}ab&amp;space;-&amp;space;b&amp;space;+&amp;space;\\frac{1}{2}a\" alt=\"\\frac{1}{16}a^2 + \\frac{1}{4}b^2 +1 - \\frac{1}{4}ab - b + \\frac{1}{2}a\" align=\"absmiddle\" \/><\/p>\n<p><span style=\"color: #000000;\"><strong>5. Factorize:<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong>(i) 4x<sup>2 <\/sup>+ 9y<sup>2 <\/sup>+ 16z<sup>2 <\/sup>+ 12xy \u2013 24yz \u2013 16xz<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong>(ii) 2x<sup>2 <\/sup>+ y<sup>2 <\/sup>+ 8z<sup>2 <\/sup>\u2013 2\u221a2xy + 4\u221a2yz \u2013 8xz<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<\/strong><br \/>\n<\/span><span style=\"color: #000000;\"><strong>(i) 4x<sup>2 <\/sup>+ 9y<sup>2 <\/sup>+ 16z<sup>2 <\/sup>+ 12xy \u2013 24yz \u2013 16xz<br \/>\n<\/strong><\/span><span style=\"color: #000000;\">Using identity, (x + y + z)<sup>2<\/sup>\u00a0= x<sup>2 <\/sup>+ y<sup>2 <\/sup>+ z<sup>2 <\/sup>+ 2xy + 2yz + 2zx<br \/>\n<\/span><span style=\"color: #000000;\">We can say that, x<sup>2 <\/sup>+ y<sup>2 <\/sup>+ z<sup>2 <\/sup>+ 2xy + 2yz + 2zx = (x + y + z)<sup>2<br \/>\n<\/sup><\/span><span style=\"color: #000000;\">4x<sup>2 <\/sup>+ 9y<sup>2 <\/sup>+ 16z<sup>2 <\/sup>+ 12xy \u2013 24yz \u2013 16xz<br \/>\n= (2x)<sup>2 <\/sup>+ (3y)<sup>2 <\/sup>+ (\u22124z)<sup>2 <\/sup>+ (2 \u00d7 2x \u00d7 3y) + (2 \u00d7 3y \u00d7 \u22124z) + (2 \u00d7 \u22124z \u00d7 2x)<br \/>\n<\/span><span style=\"color: #000000;\">= (2x + 3y \u2013 4z)<sup>2<br \/>\n<\/sup><\/span><span style=\"color: #000000;\">= (2x + 3y \u2013 4z) (2x + 3y \u2013 4z)<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(ii) 2x<sup>2 <\/sup>+ y<sup>2 <\/sup>+ 8z<sup>2 <\/sup>\u2013 2\u221a2xy + 4\u221a2yz \u2013 8xz<br \/>\n<\/strong><\/span><span style=\"color: #000000;\">Using identity, (x + y + z)<sup>2<\/sup>\u00a0= x<sup>2 <\/sup>+ y<sup>2 <\/sup>+ z<sup>2 <\/sup>+ 2xy + 2yz + 2zx<br \/>\n<\/span><span style=\"color: #000000;\">We can say that, x<sup>2 <\/sup>+ y<sup>2 <\/sup>+ z<sup>2 <\/sup>+ 2xy + 2yz + 2zx = (x + y + z)<sup>2<br \/>\n<\/sup><\/span><span style=\"color: #000000;\">2x<sup>2 <\/sup>+ y<sup>2 <\/sup>+ 8z<sup>2 <\/sup>\u2013 2\u221a2xy + 4\u221a2yz \u2013 8xz<br \/>\n<\/span><span style=\"color: #000000;\">= (-\u221a2x)<sup>2 <\/sup>+ (y)<sup>2 <\/sup>+ (2\u221a2z)<sup>2 <\/sup>+ (2 \u00d7 -\u221a2x \u00d7 y)+(2 \u00d7 y \u00d7 2\u221a2z) + (2 \u00d7 2\u221a2 \u00d7 \u2212\u221a2x)<br \/>\n<\/span><span style=\"color: #000000;\">= (\u2212\u221a2x + y + 2\u221a2z)<sup>2<br \/>\n<\/sup><\/span><span style=\"color: #000000;\">= (\u2212\u221a2x + y + 2\u221a2z) (\u2212\u221a2x + y + 2\u221a2z)<\/span><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6. Write the following cubes in expanded form:<br \/>\n<\/strong><strong>(i) (2x + 1)<sup>3<br \/>\n<\/sup><\/strong><strong>(ii) (2a \u2212 3b)<sup>3<br \/>\n<\/sup><\/strong><strong>(iii) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\left&amp;space;[&amp;space;\\frac{3}{2}x&amp;space;+&amp;space;1\\right&amp;space;]^3}\" alt=\"\\mathbf{\\left [ \\frac{3}{2}x + 1\\right ]^3}\" align=\"absmiddle\" \/><sup><br \/>\n<\/sup><\/strong><strong>(iv) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\left&amp;space;[&amp;space;x-\\frac{2}{3}y&amp;space;\\right&amp;space;]^3}\" alt=\"\\mathbf{\\left [ x-\\frac{2}{3}y \\right ]^3}\" align=\"absmiddle\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<\/strong><br \/>\n<strong>(i) (2x + 1)<sup>3<br \/>\n<\/sup><\/strong>Using identity, (x + y)<sup>3<\/sup>\u00a0= x<sup>3 <\/sup>+ y<sup>3 <\/sup>+ 3xy(x + y)<br \/>\n(2x + 1)<sup>3 <\/sup>= (2x)<sup>3 <\/sup>+ 1<sup>3 <\/sup>+ (3 \u00d7 2x \u00d7 1)(2x + 1)<br \/>\n= 8x<sup>3 <\/sup>+ 1 + 6x(2x + 1)<br \/>\n= 8x<sup>3 <\/sup>+ 12x<sup>2 <\/sup>+ 6x + 1<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) (2a \u2212 3b)<sup>3<br \/>\n<\/sup><\/strong>Using identity,(x \u2013 y)<sup>3<\/sup>\u00a0= x<sup>3 <\/sup>\u2013 y<sup>3 <\/sup>\u2013 3xy(x \u2013 y)<br \/>\n(2a \u2212 3b)<sup>3\u00a0<\/sup>= (2a)<sup>3 <\/sup>\u2212 (3b)<sup>3 <\/sup>\u2013 (3 \u00d7 2a \u00d7 3b)(2a \u2013 3b)<br \/>\n= 8a<sup>3 <\/sup>\u2013 27b<sup>3 <\/sup>\u2013 18ab(2a \u2013 3b)<br \/>\n= 8a<sup>3 <\/sup>\u2013 27b<sup>3 <\/sup>\u2013 36a<sup>2<\/sup>b + 54ab<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) <img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\left&amp;space;[&amp;space;\\frac{3}{2}x&amp;space;+&amp;space;1\\right&amp;space;]^3}\" alt=\"\\mathbf{\\left [ \\frac{3}{2}x + 1\\right ]^3}\" width=\"80\" height=\"47\" align=\"absmiddle\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">Using identity, (x + y)<sup>3<\/sup>\u00a0= x<sup>3 <\/sup>+ y<sup>3 <\/sup>+ 3xy(x + y)<br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;[&amp;space;\\frac{3}{2}x&amp;space;+&amp;space;1\\right&amp;space;]^3\" alt=\"\\left [ \\frac{3}{2}x + 1\\right ]^3\" align=\"absmiddle\" \/> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(&amp;space;\\frac{3}{2}x&amp;space;\\right&amp;space;)^3\" alt=\"\\left ( \\frac{3}{2}x \\right )^3\" align=\"absmiddle\" \/> + 1<sup>3 <\/sup>+ (3\u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{3}{2}x\" alt=\"\\frac{3}{2}x\" align=\"absmiddle\" \/> \u00d7 1)(<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{3}{2}x\" alt=\"\\frac{3}{2}x\" align=\"absmiddle\" \/> + 1) <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">= <img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{27}{8}x^3&amp;space;+&amp;space;1&amp;space;+&amp;space;\\frac{9}{2}x&amp;space;\\left&amp;space;(&amp;space;\\frac{3}{2}x&amp;space;+1&amp;space;\\right&amp;space;)\" alt=\"\\frac{27}{8}x^3 + 1 + \\frac{9}{2}x \\left ( \\frac{3}{2}x +1 \\right )\" width=\"193\" height=\"44\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{27}{8}x^3&amp;space;+&amp;space;1&amp;space;+&amp;space;\\frac{27}{4}x^2&amp;space;+&amp;space;\\frac{9}{2}x\" alt=\"\\frac{27}{8}x^3 + 1 + \\frac{27}{4}x^2 + \\frac{9}{2}x\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{27}{8}x^3&amp;space;+&amp;space;\\frac{27}{4}x^2&amp;space;+&amp;space;\\frac{9}{2}x&amp;space;+&amp;space;1\" alt=\"\\frac{27}{8}x^3 + \\frac{27}{4}x^2 + \\frac{9}{2}x + 1\" align=\"absmiddle\" \/><br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv) <img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\left&amp;space;[&amp;space;x-\\frac{2}{3}y&amp;space;\\right&amp;space;]^3}\" alt=\"\\mathbf{\\left [ x-\\frac{2}{3}y \\right ]^3}\" width=\"81\" height=\"47\" align=\"absmiddle\" \/><br \/>\n<\/strong>Using identity, (x \u2013 y)<sup>3<\/sup>\u00a0= x<sup>3 <\/sup>\u2013 y<sup>3 <\/sup>\u2013 3xy(x \u2013 y)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"> <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;[&amp;space;x-\\frac{2}{3}y&amp;space;\\right&amp;space;]^3\" alt=\"\\left [ x-\\frac{2}{3}y \\right ]^3\" align=\"absmiddle\" \/> = x<sup>3\u00a0<\/sup>&#8211; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(&amp;space;\\frac{2}{3}y&amp;space;\\right&amp;space;)^3\" alt=\"\\left ( \\frac{2}{3}y \\right )^3\" align=\"absmiddle\" \/> &#8211; (3 \u00d7 x \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{2}{3}y\" alt=\"\\frac{2}{3}y\" align=\"absmiddle\" \/> ) (x &#8211; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{2}{3}y\" alt=\"\\frac{2}{3}y\" align=\"absmiddle\" \/> )<br \/>\n<\/span><\/p>\n<p><span style=\"color: #000000;\">= x<sup>3\u00a0<\/sup>&#8211; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{8}{27}y^3\" alt=\"\\frac{8}{27}y^3\" align=\"absmiddle\" \/> &#8211; 2xy (x &#8211; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{2}{3}y\" alt=\"\\frac{2}{3}y\" align=\"absmiddle\" \/> )<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">= x<sup>3\u00a0<\/sup>&#8211; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{8}{27}y^3\" alt=\"\\frac{8}{27}y^3\" align=\"absmiddle\" \/> &#8211; 2x<sup>2<\/sup>y + <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{4}{3}xy^2\" alt=\"\\frac{4}{3}xy^2\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>7. Evaluate the following using suitable identities:<br \/>\n<\/strong><strong>(i) (99)<sup>3<br \/>\n<\/sup><\/strong><strong>(ii) (102)<sup>3<br \/>\n<\/sup><\/strong><strong>(iii) (998)<sup>3<\/sup><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<\/strong> <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) (99)<sup>3<br \/>\n<\/sup><\/strong>We have, 99 = (100 -1)<br \/>\n\u2234 99<sup>3<\/sup>\u00a0= (100 \u2013 1)<sup>3<br \/>\n<\/sup>= (100)<sup>3<\/sup>\u00a0\u2013 1<sup>3<\/sup>\u00a0\u2013 3(100)(1)(100 -1)<br \/>\n[Using (a \u2013 b)<sup>3<\/sup>\u00a0= a<sup>3<\/sup>\u00a0\u2013 b<sup>3<\/sup>\u00a0\u2013 3ab (a \u2013 b)]<br \/>\n= 1000000 \u2013 1 \u2013 300(100 \u2013 1)<br \/>\n= 1000000 -1 \u2013 30000 + 300<br \/>\n= 1000300 \u2013 30001<br \/>\n= 970299<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) (102)<sup>3<br \/>\n<\/sup><\/strong>We have, 102 =100 + 2<br \/>\n\u2234 102<sup>3<\/sup>\u00a0= (100 + 2)<sup>3<br \/>\n<\/sup>= (100)<sup>3<\/sup>\u00a0+ (2)<sup>3<\/sup> + 3(100)(2)(100 + 2)<br \/>\n[Using (a + b)<sup>3<\/sup>\u00a0= a<sup>3<\/sup>\u00a0+ b<sup>3<\/sup> + 3ab (a + b)]<br \/>\n= 1000000 + 8 + 600(100 + 2)<br \/>\n= 1000000 + 8 + 60000 + 1200<br \/>\n= 1061208<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) (998)<sup>3<br \/>\n<\/sup><\/strong>We have, 998 = 1000 \u2013 2<br \/>\n\u2234 (998)<sup>3<\/sup>\u00a0= (1000-2)<sup>3<\/sup><br \/>\n= (1000)<sup>3<\/sup>\u2013 (2)<sup>3<\/sup>\u00a0\u2013 3(1000)(2)(1000 \u2013 2)<br \/>\n[Using (a \u2013 b)<sup>3<\/sup>\u00a0= a<sup>3<\/sup>\u00a0\u2013 b<sup>3<\/sup>\u00a0\u2013 3ab (a \u2013 b)]<br \/>\n= 1000000000 \u2013 8 \u2013 6000(1000 \u2013 2)<br \/>\n= 1000000000 \u2013 8 \u2013 6000000 + 12000<br \/>\n= 994011992<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>8. Factorise each of the following:<br \/>\n<\/strong><strong>(i) 8a<sup>3 <\/sup>+ b<sup>3 <\/sup>+ 12a<sup>2<\/sup>b + 6ab<sup>2<br \/>\n<\/sup><\/strong><strong>(ii) 8a<sup>3 <\/sup>\u2013 b<sup>3 <\/sup>\u2013 12a<sup>2<\/sup>b + 6ab<sup>2<br \/>\n<\/sup><\/strong><strong>(iii) 27 \u2013 125a<sup>3 <\/sup>\u2013 135a + 225a<sup>2<\/sup><br \/>\n<\/strong><strong>(iv) 64a<sup>3 <\/sup>\u2013 27b<sup>3 <\/sup>\u2013 144a<sup>2<\/sup>b + 108ab<sup>2<br \/>\n<\/sup><\/strong><strong>(v) 27p<sup>3 <\/sup>\u2013 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{1}{216}}\" alt=\"\\mathbf{\\frac{1}{216}}\" align=\"absmiddle\" \/> \u2212 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{9}{2}}\" alt=\"\\mathbf{\\frac{9}{2}}\" align=\"absmiddle\" \/>p<sup>2 <\/sup>+ <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{1}{4}}\" alt=\"\\mathbf{\\frac{1}{4}}\" align=\"absmiddle\" \/>p<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<\/strong> <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) 8a<sup>3 <\/sup>+ b<sup>3 <\/sup>+ 12a<sup>2<\/sup>b + 6ab<sup>2<br \/>\n<\/sup><\/strong>= (2a)<sup>3<\/sup>\u00a0+ (b)<sup>3<\/sup>\u00a0+ 6ab(2a + b)<br \/>\n= (2a)<sup>3<\/sup>\u00a0+ (b)<sup>3<\/sup>\u00a0+ 3(2a)(b)(2a + b)<br \/>\n= (2a + b)<sup>3\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 <\/sup>[Using a<sup>3<\/sup>\u00a0+ b<sup>3<\/sup>\u00a0+ 3 ab(a + b) = (a + b)<sup>3<\/sup>]<br \/>\n= (2a + b)(2a + b)(2a + b)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) 8a<sup>3 <\/sup>\u2013 b<sup>3 <\/sup>\u2013 12a<sup>2<\/sup>b + 6ab<sup>2<br \/>\n<\/sup><\/strong>= (2a)<sup>3<\/sup>\u00a0\u2013 (b)<sup>3<\/sup> \u2013 3(2a)(b)(2a \u2013 b)<br \/>\n= (2a \u2013 b)<sup>3\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 <\/sup>[Using a<sup>3<\/sup>\u00a0+ b<sup>3<\/sup>\u00a0+ 3 ab(a + b) = (a + b)<sup>3<\/sup>]<br \/>\n= (2a \u2013 b)(2a \u2013 b)(2a \u2013 b)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) 27 \u2013 125a<sup>3 <\/sup>\u2013 135a + 225a<sup>2<\/sup><br \/>\n<\/strong>= (3)<sup>3<\/sup>\u00a0\u2013 (5a)<sup>3<\/sup> \u2013 3(3)(5a)(3 \u2013 5a)<br \/>\n= (3 \u2013 5a)<sup>3\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 <\/sup>[Using a<sup>3<\/sup>\u00a0+ b<sup>3<\/sup>\u00a0+ 3 ab(a + b) = (a + b)<sup>3<\/sup>]<br \/>\n= (3 \u2013 5a)(3 \u2013 5a)(3 \u2013 5a)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv) 64a<sup>3 <\/sup>\u2013 27b<sup>3 <\/sup>\u2013 144a<sup>2<\/sup>b + 108ab<sup>2<br \/>\n<\/sup><\/strong>= (4a)<sup>3<\/sup>\u00a0\u2013 (3b)<sup>3<\/sup> \u2013 3(4a)(3b)(4a \u2013 3b)<br \/>\n= (4a \u2013 3b)<sup>3\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 <\/sup>[Using a<sup>3<\/sup>\u00a0\u2013 b<sup>3<\/sup>\u00a0\u2013 3 ab(a \u2013 b) = (a \u2013 b)<sup>3<\/sup>]<br \/>\n= (4a \u2013 3b)(4a \u2013 3b)(4a \u2013 3b)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(v) 27p<sup>3 <\/sup>\u2013 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{1}{216}}\" alt=\"\\mathbf{\\frac{1}{216}}\" align=\"absmiddle\" \/> \u2212 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{9}{2}}\" alt=\"\\mathbf{\\frac{9}{2}}\" align=\"absmiddle\" \/>p<sup>2 <\/sup>+ <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{1}{4}}\" alt=\"\\mathbf{\\frac{1}{4}}\" align=\"absmiddle\" \/>p <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">= (3p)<sup>3 <\/sup>\u2013 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(&amp;space;\\frac{1}{6}&amp;space;\\right&amp;space;)^3\" alt=\"\\left ( \\frac{1}{6} \\right )^3\" align=\"absmiddle\" \/> \u2212 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(&amp;space;\\frac{9}{2}&amp;space;\\right&amp;space;)\" alt=\"\\left ( \\frac{9}{2} \\right )\" align=\"absmiddle\" \/>p<sup>2 <\/sup>+ <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{4}\" alt=\"\\frac{1}{4}\" align=\"absmiddle\" \/>p <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">= (3p)<sup>3 <\/sup>\u2013 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(&amp;space;\\frac{1}{6}&amp;space;\\right&amp;space;)^3\" alt=\"\\left ( \\frac{1}{6} \\right )^3\" align=\"absmiddle\" \/> \u2212 3(3p\u00d7<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{6}\" alt=\"\\frac{1}{6}\" align=\"absmiddle\" \/>)(3p \u2013 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{6}\" alt=\"\\frac{1}{6}\" align=\"absmiddle\" \/>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Using (x \u2013 y)<sup>3<\/sup>\u00a0= x<sup>3<\/sup>\u00a0\u2013 y<sup>3<\/sup> \u2013 3xy (x \u2013 y)<br \/>\nTaking x = 3p and y = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{6}\" alt=\"\\frac{1}{6}\" align=\"absmiddle\" \/> <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">= (3p \u2013 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{6}\" alt=\"\\frac{1}{6}\" align=\"absmiddle\" \/>)<sup>3<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">= (3p \u2013 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{6}\" alt=\"\\frac{1}{6}\" align=\"absmiddle\" \/>)(3p \u2013 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{6}\" alt=\"\\frac{1}{6}\" align=\"absmiddle\" \/>)(3p \u2013 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{6}\" alt=\"\\frac{1}{6}\" align=\"absmiddle\" \/>)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>9. Verify:<br \/>\n<\/strong><strong>(i) x<sup>3 <\/sup>+ y<sup>3\u00a0<\/sup>= (x + y)(x<sup>2 <\/sup>\u2013 xy + y<sup>2<\/sup>)<br \/>\n<\/strong><strong>(ii) x<sup>3 <\/sup>\u2013 y<sup>3\u00a0<\/sup>= (x \u2013 y)(x<sup>2 <\/sup>+ xy + y<sup>2<\/sup>)<br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<\/strong><br \/>\n<strong>(i) x<sup>3 <\/sup>+ y<sup>3\u00a0<\/sup>= (x + y)(x<sup>2 <\/sup>\u2013 xy + y<sup>2<\/sup>)<br \/>\n<\/strong>We know that, (x + y)<sup>3<\/sup>\u00a0= x<sup>3 <\/sup>+ y<sup>3 <\/sup>+ 3xy(x + y)<br \/>\n\u21d2 x<sup>3 <\/sup>+ y<sup>3\u00a0<\/sup>= (x + y)<sup>3 <\/sup>\u2013 3xy(x + y)<br \/>\n\u21d2 x<sup>3 <\/sup>+ y<sup>3\u00a0<\/sup>= (x+y)[(x + y)<sup>2<\/sup>\u20133xy]<br \/>\nTaking (x + y) common<br \/>\n\u21d2 x<sup>3 <\/sup>+ y<sup>3\u00a0<\/sup>= (x + y)[(x<sup>2 <\/sup>+ y<sup>2 <\/sup>+ 2xy) \u2013 3xy]<br \/>\n\u21d2 x<sup>3 <\/sup>+ y<sup>3\u00a0<\/sup>= (x + y)(x<sup>2 <\/sup>+ y<sup>2 <\/sup>\u2013 xy)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) x<sup>3 <\/sup>\u2013 y<sup>3\u00a0<\/sup>= (x \u2013 y)(x<sup>2 <\/sup>+ xy + y<sup>2<\/sup>)<br \/>\n<\/strong>We know that, (x \u2013 y)<sup>3<\/sup>\u00a0= x<sup>3 <\/sup>\u2013 y<sup>3 <\/sup>\u2013 3xy(x \u2013 y)<br \/>\n\u21d2 x<sup>3 <\/sup>\u2212 y<sup>3\u00a0<\/sup>= (x \u2013 y)<sup>3 <\/sup>+ 3xy(x \u2013 y)<br \/>\n\u21d2 x<sup>3 <\/sup>\u2212 y<sup>3\u00a0<\/sup>= (x \u2013 y)[(x \u2013 y)<sup>2 <\/sup>+ 3xy]<br \/>\nTaking (x + y) common<br \/>\n\u21d2 x<sup>3 <\/sup>\u2212 y<sup>3\u00a0<\/sup>= (x \u2013 y)[(x<sup>2 <\/sup>+ y<sup>2 <\/sup>\u2013 2xy) + 3xy]<br \/>\n\u21d2 x<sup>3 <\/sup>+ y<sup>3\u00a0<\/sup>= (x \u2013 y)(x<sup>2 <\/sup>+ y<sup>2 <\/sup>+ xy)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>10. Factorize each of the following:<br \/>\n<\/strong><strong>(i) 27y<sup>3 <\/sup>+ 125z<sup>3<br \/>\n<\/sup><\/strong><strong>(ii) 64m<sup>3 <\/sup>\u2013 343n<sup>3<\/sup><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<\/strong><br \/>\n<strong>(i) 27y<sup>3 <\/sup>+ 125z<sup>3<br \/>\n<\/sup><\/strong>We have, 27y<sup>3<\/sup>\u00a0+ 125z<sup>3<\/sup>\u00a0= (3y)<sup>3<\/sup>\u00a0+ (5z)<sup>3<\/sup><br \/>\nWe know that, x<sup>3 <\/sup>+ y<sup>3\u00a0<\/sup>= (x + y)(x<sup>2 <\/sup>\u2013 xy + y<sup>2<\/sup>)<br \/>\n27y<sup>3 <\/sup>+ 125z<sup>3\u00a0<\/sup>= (3y)<sup>3 <\/sup>+ (5z)<sup>3<br \/>\n<\/sup>= (3y + 5z)[(3y)<sup>2 <\/sup>\u2013 (3y)(5z) + (5z)<sup>2<\/sup>]<br \/>\n= (3y + 5z)(9y<sup>2 <\/sup>\u2013 15yz + 25z<sup>2<\/sup>)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) 64m<sup>3<\/sup>\u2013343n<sup>3<br \/>\n<\/sup><\/strong>We have, 64m<sup>3<\/sup>\u00a0\u2013 343n<sup>3<\/sup>\u00a0= (4m)<sup>3<\/sup>\u00a0\u2013 (7n)<sup>3<br \/>\n<\/sup>We know that, x<sup>3 <\/sup>\u2013 y<sup>3 <\/sup>= (x \u2013 y)(x<sup>2 <\/sup>+ xy + y<sup>2<\/sup>)<br \/>\n64m<sup>3 <\/sup>\u2013 343n<sup>3\u00a0<\/sup>= (4m)<sup>3 <\/sup>\u2013 (7n)<sup>3<br \/>\n<\/sup>= (4m \u2013 7n)[(4m)<sup>2 <\/sup>+ (4m)(7n) + (7n)<sup>2<\/sup>]<br \/>\n= (4m \u2013 7n)(16m<sup>2 <\/sup>+ 28mn + 49n<sup>2<\/sup>)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>11. Factorise : 27x<sup>3 <\/sup>+ y<sup>3 <\/sup>+ z<sup>3 <\/sup>\u2013 9xyz\u00a0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<\/strong><br \/>\nWe have, 27x<sup>3<\/sup>\u00a0+ y<sup>3<\/sup>\u00a0+ z<sup>3<\/sup>\u00a0\u2013 9xyz = (3x)<sup>3<\/sup>\u00a0+ (y)<sup>3<\/sup>\u00a0+ (z)<sup>3<\/sup>\u00a0\u2013 3(3x)(y)(z)<br \/>\nUsing the identity, x<sup>3<\/sup>\u00a0+ y<sup>3<\/sup>\u00a0+ z<sup>3<\/sup>\u00a0\u2013 3xyz = (x + y + z)(x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0+ z<sup>2<\/sup>\u00a0\u2013 xy \u2013 yz \u2013 zx)<br \/>\nWe have, (3x)<sup>3<\/sup>\u00a0+ (y)<sup>3<\/sup>\u00a0+ (z)<sup>3<\/sup> \u2013 3(3x)(y)(z)<br \/>\n= (3x + y + z)[(3x)<sup>3<\/sup>\u00a0+ y<sup>3<\/sup>\u00a0+ z<sup>3<\/sup>\u00a0\u2013 (3x \u00d7 y) \u2013 (y \u00d7 2) \u2013 (z \u00d7 3x)]<br \/>\n= (3x + y + z)(9x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0+ z<sup>2<\/sup>\u00a0\u2013 3xy \u2013 yz \u2013 3zx)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>12. Verify that: <\/strong><strong>x<sup>3 <\/sup>+ y<sup>3 <\/sup>+ z<sup>3 <\/sup>\u2013 3xyz = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{1}{2}}\" alt=\"\\mathbf{\\frac{1}{2}}\" align=\"absmiddle\" \/>(x + y + z)[(x \u2013 y)<sup>2 <\/sup>+ (y \u2013 z)<sup>2 <\/sup>+ (z \u2013 x)<sup>2<\/sup>] <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<\/strong>\u00a0 \u00a0We know that, x<sup>3 <\/sup>+ y<sup>3 <\/sup>+ z<sup>3 <\/sup>\u2212 3xyz = (x + y + z)(x<sup>2 <\/sup>+ y<sup>2 <\/sup>+ z<sup>2 <\/sup>\u2013 xy \u2013 yz \u2013 xz)<br \/>\n\u21d2 x<sup>3 <\/sup>+ y<sup>3 <\/sup>+ z<sup>3 <\/sup>\u2013 3xyz = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/>(x + y + z)[2(x<sup>2 <\/sup>+ y<sup>2 <\/sup>+ z<sup>2 <\/sup>\u2013 xy \u2013 yz \u2013 xz)]<br \/>\n= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> (x + y + z)(2x<sup>2 <\/sup>+ 2y<sup>2 <\/sup>+ 2z<sup>2 <\/sup>\u2013 2xy \u2013 2yz \u2013 2xz)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/>(x + y + z)[(x<sup>2 <\/sup>+ y<sup>2 <\/sup>\u2212 2xy) + (y<sup>2 <\/sup>+ z<sup>2 <\/sup>\u2013 2yz) + (x<sup>2 <\/sup>+ z<sup>2 <\/sup>\u2013 2xz)]<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/>(x + y + z)[(x \u2013 y)<sup>2 <\/sup>+ (y \u2013 z)<sup>2 <\/sup>+ (z \u2013 x)<sup>2<\/sup>]<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>13. If\u00a0 x + y + z = 0, show that x<sup>3 <\/sup>+ y<sup>3 <\/sup>+ z<sup>3\u00a0<\/sup>= 3xyz.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<\/strong> Since, x + y + z = 0<br \/>\n\u21d2 x + y = -z<br \/>\n\u21d2 (x + y)<sup>3<\/sup>\u00a0= (-z)<sup>3<\/sup><br \/>\n\u21d2 x<sup>3<\/sup>\u00a0+ y<sup>3<\/sup>\u00a0+ 3xy(x + y) = -z<sup>3<\/sup><br \/>\n\u21d2 x<sup>3<\/sup>\u00a0+ y<sup>3<\/sup>\u00a0+ 3xy(-z) = -z<sup>3<\/sup>\u00a0[\u2235 x + y = -z]<br \/>\n\u21d2 x<sup>3<\/sup>\u00a0+ y<sup>3<\/sup>\u00a0\u2013 3xyz = -z<sup>3<\/sup><br \/>\n\u21d2 x<sup>3<\/sup>\u00a0+ y<sup>3<\/sup>\u00a0+ z<sup>3<\/sup>\u00a0= 3xyz<br \/>\nHence, if x + y + z = 0, then<br \/>\nx<sup>3<\/sup>\u00a0+ y<sup>3<\/sup>\u00a0+ z<sup>3<\/sup>\u00a0= 3xyz<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>14. Without actually calculating the cubes, find the value of each of the following:<br \/>\n<\/strong><strong>(i) (\u221212)<sup>3 <\/sup>+ (7)<sup>3 <\/sup>+ (5)<sup>3<br \/>\n<\/sup><\/strong><strong>(ii) (28)<sup>3 <\/sup>+ (\u221215)<sup>3 <\/sup>+ (\u221213)<sup>3<\/sup><br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<\/strong> <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) (\u221212)<sup>3 <\/sup>+ (7)<sup>3 <\/sup>+ (5)<sup>3<br \/>\n<\/sup><\/strong>Let x = \u221212, y = 7 and z= 5<br \/>\nWe know that if x + y + z = 0, then x<sup>3 <\/sup>+ y<sup>3 <\/sup>+ z<sup>3 <\/sup>= 3xyz.<br \/>\nHere, \u221212 + 7 + 5 = 0<br \/>\n(\u221212)<sup>3 <\/sup>+ (7)<sup>3 <\/sup>+ (5)<sup>3\u00a0<\/sup>= 3xyz<br \/>\n= 3 \u00d7 -12 \u00d7 7 \u00d7 5<br \/>\n= -1260<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) (28)<sup>3 <\/sup>+ (\u221215)<sup>3 <\/sup>+ (\u221213)<sup>3<br \/>\n<\/sup><\/strong>Let x = 28,\u00a0y= \u221215 and\u00a0z = \u221213<br \/>\nWe know that if x + y + z = 0, then x<sup>3 <\/sup>+ y<sup>3 <\/sup>+ z<sup>3\u00a0<\/sup>= 3xyz.<br \/>\nHere, x + y + z = 28 \u2013 15 \u2013 13 = 0<br \/>\n(28)<sup>3 <\/sup>+ (\u221215)<sup>3 <\/sup>+ (\u221213)<sup>3\u00a0<\/sup>= 3xyz<br \/>\n= 0 + 3(28)(\u221215)(\u221213)<br \/>\n= 16380<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:<br \/>\n<\/strong><strong>(i) Area :\u00a025a<sup>2 <\/sup>\u2013 35a + 12<br \/>\n<\/strong><strong>(ii) Area :\u00a035y<sup>2 <\/sup>+ 13y \u2013 12<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<\/strong><br \/>\n<strong>Area of a rectangle = (Length) x (Breadth)<br \/>\n<\/strong><strong>(i) Area :\u00a025a<sup>2 <\/sup>\u2013 35a + 12<br \/>\n<\/strong>25a<sup>2<\/sup>\u00a0\u2013 35a + 12<br \/>\n= 25a<sup>2<\/sup>\u00a0\u2013 20a \u2013 15a + 12<br \/>\n= 5a(5a \u2013 4) \u2013 3(5a \u2013 4)<br \/>\n= (5a \u2013 4)(5a \u2013 3)<br \/>\nPossible expression for length\u00a0 = 5a\u20134<\/span><br \/>\n<span style=\"color: #000000;\">Possible expression for breadth\u00a0 = 5a \u20133<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) Area :\u00a035y<sup>2 <\/sup>+ 13y \u2013 12<br \/>\n<\/strong>35y<sup>2<\/sup>+ 13y -12<br \/>\n= 35y<sup>2<\/sup> + 28y \u2013 15y &#8211; 12<br \/>\n= 7y(5y + 4) \u2013 3(5y + 4)<br \/>\n= (5y + 4)(7y \u2013 3)<br \/>\nPossible expression for length\u00a0 =\u00a0(5y+4)<br \/>\nPossible expression for breadth\u00a0 =\u00a0(7y\u20133)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?<br \/>\n<\/strong><strong>(i) Volume :\u00a03x<sup>2 <\/sup>\u2013 12x<br \/>\n<\/strong><strong>(ii) Volume :\u00a012ky<sup>2 <\/sup>+ 8ky \u2013 20k<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<\/strong><br \/>\n<strong>Volume of a cuboid = (Length) x (Breadth) x (Height)<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) Volume :\u00a03x<sup>2<\/sup>\u201312x<br \/>\n<\/strong>We have, 3x<sup>2<\/sup>\u00a0\u2013 12x = 3(x<sup>2<\/sup> \u2013 4x)<br \/>\n= 3 \u00d7 <em>x<\/em> \u00d7 (<em>x<\/em> \u2013 4)<br \/>\nPossible expression for length\u00a0=\u00a03<br \/>\nPossible expression for breadth\u00a0=\u00a0<em>x<\/em><br \/>\nPossible expression for height\u00a0=\u00a0(<em>x <\/em>\u2013 4)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) Volume :\u00a012ky<sup>2 <\/sup>+ 8ky \u2013 20k<br \/>\n<\/strong>We have, 12ky<sup>2<\/sup>\u00a0+ 8ky \u2013 20k<br \/>\n= 4[3ky<sup>2<\/sup>\u00a0+ 2ky \u2013 5k]<br \/>\n= 4[k(3y<sup>2<\/sup> + 2y \u2013 5)]<br \/>\n= 4 \u00d7 k \u00d7 (3y<sup>2<\/sup>\u00a0+ 2y \u2013 5)<br \/>\n= 4k[3y<sup>2<\/sup>\u00a0\u2013 3y + 5y \u2013 5]<br \/>\n= 4k[3y(y \u2013 1) + 5(y \u2013 1)]<br \/>\n= 4k[(3y + 5) \u00d7 (y \u2013 1)]<br \/>\n= 4k \u00d7 (3y + 5) \u00d7 (y \u2013 1)<br \/>\nPossible expression for length\u00a0= 4k<br \/>\nPossible expression for breadth\u00a0=\u00a0(3y +5)<br \/>\nPossible expression for height = (y &#8211; 1)<\/span><\/p>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 9 Maths\u00a0 Chapter &#8211; 2 (Polynomials)\u00a0 The NCERT Solutions in English Language for Class 9 Mathematics\u00a0Chapter &#8211; 2 Polynomials Exercise 2.5 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter 2: Polynomials\u00a0 NCERT Solution Class 9 Maths Ex &#8211; 2.1 NCERT Solution Class 9 Maths [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[698],"tags":[703,702,186,704,701,5],"class_list":["post-4223","post","type-post","status-publish","format-standard","hentry","category-class-9-maths","tag-class-9th-chapter-2-polynomials-in-english","tag-ncert-solution-class-9","tag-ncert-solutions","tag-ncert-solutions-class-9-maths-chapter-2-in-english","tag-ncert-solutions-class-9-maths-in-english","tag-ncert-solutions-in-english"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.4) - 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