NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3

NCERT Solutions Class 8 Mathematics 
Chapter – 9 (Algebraic Expressions and Identities) 

The NCERT Solutions in English Language for Class 8 Mathematics Chapter – 9 Algebraic Expressions and Identities Exercise 9.3 has been provided here to help the students in solving the questions from this exercise. 

Chapter 9: Algebraic Expressions and Identities

• NCERT Solution Class 8 Maths Ex – 9.1
• NCERT Solution Class 8 Maths Ex – 9.2
• NCERT Solution Class 8 Maths Ex – 9.4
• NCERT Solution Class 8 Maths Ex – 9.5

Exercise – 9.3 

1. Carry out the multiplication of the expressions in each of the following pairs.
(i) 4p, q + r
(ii) ab, a – b
(iii) a + b, 7a²b²
(iv) a² – 9, 4a
(v) pq + qr + rp, 0

Solution –

(i) 4p, q + r
= 4p(q + r)
= 4pq + 4pr

(ii) ab, a – b
= ab(a – b)
= a²b – ab²

(iii) a + b, 7a²b²
= (a + b) (7a²b²)
= 7a³b² + 7a²b³

(iv) a² – 9, 4a
= (a² – 9)(4a)
= 4a³ – 36a

(v) pq + qr + rp, 0
= (pq + qr + rp) × 0
= 0 ( Anything multiplied by zero is zero )

2. Complete the table.

Solution –

3. Find the product.
(i) a² × (2a²²) × (4a²⁶)
(ii)  ×

(iii) ×
(iv) (x) × (x²) × (x³) × (x⁴)

Solution –

(i) a² × (2a²²) × (4a²⁶)
= (2 × 4) ( a² × a²² × a²⁶ )
= 8 × a^{2 + 22 + 26}
= 8a⁵⁰

(ii)  ×

= ( x × x² × y × y² )

= x³y³

(iii) ×

= (p × p³× q³ × q)
= – 4p⁴q⁴

(iv) (x) × (x²) × (x³) × (x⁴)
= x ^{1 + 2 + 3 + 4}
=  x¹⁰

4.
(a) Simplify 3x (4x – 5) + 3 and find its values for
(i) x = 3
(ii) x =
(b) Simplify a (a²+ a + 1) + 5 and find its value for
(i) a = 0,
(ii) a = 1
(iii) a = – 1

Solution –

(a) Simplify 3x (4x – 5) + 3 and find its values for
= 3x (4x – 5) + 3
= 3x (4x) – 3x ( 5) + 3
= 12x² – 15x + 3
(i) For x = 3,
= 12x² -15x + 3
= 12(3)² -15(3) + 3
= 108 – 45 + 3
= 66

(ii) For x =
= 12x² -15x + 3
= 12 – 15  + 3
= 3 – + 3

=  

=

(b) a(a² + a + 1) + 5
=a³ + a² + a + 5

(i) For a = 0,
a³ + a² + a + 5
= 0 + 0 + 0 + 5
= 5

(ii) For a = 1,
a³ + a² + a + 5
= (1)³ + (1)² + 1 + 5
= 1 + 1 + 1 + 5
= 8

(iii) For a = -1,
a³ + a² + a + 5
= (-1)³ + (-1)² + (-1) + 5
= – 1 + 1 -1 + 5
= 4

5.
(a) Add: p ( p – q), q ( q – r) and r ( r – p)
(b) Add: 2x (z – x – y) and 2y (z – y – x)
(c) Subtract: 3l (l – 4 m + 5 n) from 4l ( 10 n – 3 m + 2 l )
(d) Subtract: 3a (a + b + c ) – 2 b (a – b + c)  from 4c ( – a + b + c )

Solution –

(a) Add: p ( p – q), q ( q – r) and r ( r – p)
= p ( p – q) + q ( q – r) + r ( r – p)
= (p² – pq) + (q² – qr) + (r² – pr)
= p² + q² + r² – pq – qr – pr

(b) Add: 2x (z – x – y) and 2y (z – y – x)
= 2x (z – x – y) + 2y (z – y – x)
= (2xz – 2x² – 2xy) + (2yz – 2y² – 2xy)
= 2xz – 4xy + 2yz – 2x² – 2y²

(c) Subtract: 3l (l – 4 m + 5 n) from 4l ( 10 n – 3 m + 2 l )
= 4l ( 10 n – 3 m + 2 l ) – 3l (l – 4 m + 5 n)
= (40ln – 12lm + 8l²) – (3l² – 12lm + 15ln)
= 40ln – 12lm + 8l² – 3l² +12lm -15 ln
= 25 ln + 5l²

(d) Subtract: 3a (a + b + c ) – 2 b (a – b + c)  from 4c ( – a + b + c )
= 4c ( – a + b + c ) – (3a (a + b + c ) – 2 b (a – b + c))
= (-4ac + 4bc + 4c²) – (3a² + 3ab + 3ac – ( 2ab – 2b² + 2bc ))
=-4ac + 4bc + 4c² – (3a² + 3ab + 3ac – 2ab + 2b² – 2bc)
= -4ac + 4bc + 4c² – 3a² – 3ab – 3ac +2ab – 2b² + 2bc
= -7ac + 6bc + 4c² – 3a² – ab – 2b²