# OP Malhotra Class 9 ICSE Maths Solutions Chapter 3

## OP Malhotra Class 9 ICSE Maths Solutions Chapter 3 Expansions

Welcome to NCTB Solutions. Here with this post we are going to help 9th class students for the Solutions of OP Malhotra Class 9 ICSE Math Book, Chapter 3, Expansions. Here students can easily find step by step solutions of all the problems for Expansions, Exercise 3a and 3b Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 3 solutions. Here in this post all the solutions are based on ICSE latest Syllabus.

Expansions Exercise 3(a) Solution :

Question no – (1)

Solution :

(i) (x + 4) (x + 2)

= x2 + 2x + 4x + 8

= (x + 2) (x + 4)

= x2 + 6x + 8

(ii) (4a – 5) (5a + 6)

= 20a2 + 24a – 25a – 30

= 20a2 – a – 30

(iii) (xy + 6) (xy – 5)

= x2y2 – 5xy + 6xy – 30

= x2y2 – xy – 30

(iv) (7x2 – 5y) (x2 – 3y)

= 7x4 – 21x2y – 5x2y2 + 15y2

= 7x4 – 26x2y + 15y2

Question no – (2)

Solution :

(i) 3x + 5y

= (3x + 5y)2

= (3x)2 + 2.3x.5y + (5y)2

= 9x2 + 30xy + 25y2

(ii) 5y – 2z

= (5y – 2z)2

= (5y)2 – 2.5y.2z + (2z)2

= 25y2 – 20y2 + 2z2

(iii) 5p + 1/4q

= (5p – 1/4q)2

= (5p)2 – 2.5p.1/4q + (1/4q)2

= 25p2 – 5/2 p.q + 1/1622

(iv) (5x + 3y + z)2

= (5x + 3y + z)2

= {(5x +3y) + z}2

= (5x + 3y)2 + 2. (5x + 3y).z + z2

= (5x)2 + 2.5x.3y + 9y2 + 10xz + 6yz + z2

= 25x2 + 30xy + 9y2 + 10xz + 6yx + z2

Question no – (3) Simplify

Solution :

(2x – p + c)2 – (2x + p – c)2

∴ (4x2 + p2 + c2 – 4px – 2pc + 4cx) – (4x2 + p2 + c2 + 4xp – 2pc – 4cx)

= 4x2 + p2 + c2 – 4xp – 2cp – 4x2 + 4cx – p2 – c2 – 4px + 2cp + 4cx

= – 8xp + 8cx

= – 8xp + 8xc…(Simplified)

Question no – (4)

Solution :

(i) (3b + 7) (3b – 7)

= (b)2 – (7)2

= 9b– 492

Therefore, the product is 9b– 492

(ii) (1/3 – 5x) (1/3 + 5x)

= (1/3)2 – (5x)2

= 1/9 – 25x2

So, the product is 1/9 – 25x2

(iii) (x3 – 3) (x3 + 3)

= (x3 – 3) – (3)2

= x6 – 9

Hence, the product is x6 – 9

(iv) (a4 – 1/5y) (a4 + 1/5y)

= (a4)2 – (1/5y)2

= a8 – 1/25y2

Thus, the product is a8 – 1/25y2

Question no – (5)

Solution :

(i) (x + y) (x – y) (x2 + y2)

= (x2 – y2) (x2 + y2)

= (x2)2 – (y2)2

= x4 – y4

Thus, the product will be x4 – y4

(ii) (a2 + b2) (a4 + b4) (a + b) (a – b)

= (a2 + b2) (a4 + b4) (a2 – b2)

= {(a2)2 – (b2)2

= (a4 – b4) (a4 + b4)

= (a4)2 – (b4)2

= a8 – b8

Thus, the product will be a8 – b8

Question no – (6)

Solution :

(i) Given, x2 + 8x + 16

= x2 + 2.x4 + 42

= (x + 4)2

(It is Perfect square)

(ii) y2 + 3y + 9

= y2 + 2.y.3 + 32

(It is not perfect square)

(iii) 4m2 – 4m + 1

= (2m)2 – 2.2m.1 + 12

= (2m – 1)2

(It is perfect square)

(iv) 4x2 – 2 + 1/4x2

= (2x)2 – 2.2x, 1/2x + (1/2x)2

= (2x – 1/2x)2

(It is perfect square)

(v) m2 + 6m + 4

= m2 + 2.m.3 + 32 + 4

∴ (It is perfect square)

Question no – (7)

Solution :

Given, 4x2 – 12x + k

= (2x)2 – 2.2x × 3 + 32

Comparing we get,

k = 32

k =  9

Thus, the value of k will be 9

Question no – (8)

Solution :

(i) 4a2 + 28a

= (2a)2 + 2.2a.7a + (7)2

∴ To complete perfect square we have to add 72 – 49

∴ (2a + 7)2

(ii) 36a2 + 49b2

= (6a)2 + (7b)2 + 2.6a.7b

To complete square we have to add

= 2 × 6a × 7b = 84ab

(iii) 4a2 + 81

= (2a)2 + 2.2a.9 + 92

To complete perfect square we have to add 36a

∴ (2a + 9)2

(iv) 9a2 + 2ab + b2

= (3a)2 + 2.3a.b + b2

To complete perfect square, we have to add 6ab – 2ab = 4ab

∴ (3a + b)2

Question no – (9)

Solution :

(i) (a + 1)3

= 3a + 3.a2.1 + 3.a.12 + 13

= a3 + 3a2 + 3a + 1

(ii) (3x – 2y)3

= (3x)3 – 3. (3x)2 2y. + 3.3x (2y)2 + (2y)3

= 27x3 – 54x2y + 36xy2 + 8y3

(iii) (x2 + y)3

= (x2)3 + 3.(x2)2.y + 3.x2.y2 + (y)3

= x⁶ + 6 3x4y + 3x2y2 + y3

(iv) (2x – 1/3x)

= (2x)3 – 3. (2x)2. 1/3x.2x (1/3x)2 – (1/3x)3

= 8x– 3x 4x2, 1/3x + 6x. 1/9x2 – 1/27x3

= 8x3 – 4x + 2/3x – 1/27x3

Expansions Exercise 3(b) Solution :

Question no – (1)

Solution :

(i) a + b = 9,

ab = 20

Now,  = a2 + b2

= (a + b)2 – 2ab

= (9)2 – 2 × 20 [∵ a + b = 9, ab = 20]

= 81 – 40

= 41

So, the value of a2 + bis 41

(ii) (p + q)2 + (p – q)2 = 2, (p2 + q2)

So, 2(p2 + q2) = (p + q)2 + (p – q)2

= (14)2 + 62

= 196 + 36

= 232/2

∴ p2 + q= 16

So, the value of p2 + qis 16

(iii) Given, m + n = 8 and m – n = 2

= m + n = 8 – (i)

= m – n = 2 – (ii)

= 1 + (ii)

= 2m = 8 + 2

= m = 10/2

= 5

Putting value in (i)

= m + n = 8

= n = 8 – 5

= 3

So, mn = 5 × 3

= 15

Thus, the value of mn is 15.

(iv) x + 1/x = 3 (given)

= (x + 1/x)2 = 32

= x2 + 2.x . 1/x + 1/x2 = 9

= x2 + 1/x2 = 9 – 2

= 7

Again squaring both side,

= (x2 + 1/x2)2 = (7)2

= (x2)2 + 2.x2, 1/x2 + (1/x2)2 = 49

= x4 + 1/x4 = 49 – 2

= 47

Hence, the value is 47

(v) x + 1/x = √5

∴ x + 1/x = √5

= (x + 1/x)2 = (√5)2

= x2 + 2.x. 1/x + 1/x2 = 5 – 2

= 3

∴ (x – 1/x)2 = x2 + 1/x2 – 2

= 3 – 2

= – 1

∴ x – 1/x = ± 1

So, the value is ± 1

Question no – (2)

Solution :

(i) Given, a2 + b2 + c2 find a + b + c = 17

Squaring both side,

= (a + b + c)2 = (17)2

= a+ b2 + c2 + 2 (ab + bc + ca) = 289

= a2 + b2 + c2 + 2 × 30 = 289

= a+ b+ c2 = 289 – 60

= 229

So, the value is 229

(ii) According to the question,

a + b + c = 15

a2 + b2 + c2 = 77

Squaring both side,

= (a + b + c)2 = (15)2

= a2 + b2 + c2 + 2 (ab + bc + ca) = 225

= 77 + 2 (ab + bc + ca) = 225

= 2 (ab + bc + ca) = 225 – 77 = 148

(ab + bc + ca) = 148/2

= 74

Thus, the value of ab + bc + ca is 74

(iii) a2 + b2 + c2 = 50,

ab + bc + ca = 47

(a + b + c)2

= a2 + b2 + c2 + 2 (ab + bc + ca)

= 50 + 2 × 94

= 50 + 94

= a + b + c = √144

=  ± 12

Thus, the value of a + b + c is ± 12

Question no – (3)

Solution :

(i) Given,

x = 1,

y = 2

∴ 8x3 + 84x2y + 294xy2 + 343y3

= (2x)3 + 3. (2x)2, 7y + 3, 2x. (7y)2 + (7y)3

= (2x + 7y)3

= (2 × 1 + 7 × 2)3

= (2 + 14)3

= (16)3

= 4096

Thus, the value will be 4096

(ii) In the question, x = 2, y = 1

∴ (3x)3 – 3 (3x)3. y + 3. (3x)2.y2 – y3

= (3x – y)2

= (3 × 2 – 1)3

= (6 – 1)3

= 53

= 125

Hence, the value is 125

Question no – (4)

Solution :

(i)  a + b = 3

ab = 2

∴ a + b + 3 = ab = 2

= a+ b2 = (a + b)3 – 3ab (a + b)

= (3)3 – 3 × 2 × 3

= 27 – 18

= 9

Thus, the value of a3 + bis 9

(ii) (x + 1/x) = 3 (according to the question)

∴ x + 1/x = 3

= x3 + 1/x3 = (x + 1/x)3 – 3 x. 1/x (x + 1/x)

= (3)3 – 3 × 3

= 27 – 9

= 18

Therefore, the value is 18.

(iii) x3 – 1/x3 if x2 + 1/x2 = 18 (given)

= x2 + 1/x2 – 2 = 18 – 2

= (x – 1/x)2 = 16

= (x – 1/x) = (4)2

= (x – 1/x) = 4

x3 – 1/x3
= (x – 1/x)3 + 3. (x – 1/x)

= (4)3 + 3 × 4

= 64 + 12

= 76

Thus, the value is 76

(iv) x3 + 1/125x3 if x2 + 1/25x2 = 8 3/5 …(given)

= x2 + 1/15x2 = 8 3/5

= (x + 1/5x)2 = 43/5

= x2 + 5x2 – 2x. 1/5x = 43/5

= (x + 1/5x)= 43/5 + 2

= 45/5

Now, x3 + 1/125x3

= (x + 1/5x)3 – 3 3.x 1/5x

= (3)3 – 3/5 × 3

= 27 – 9/5

= 135 – 9/5

= 126/5

= 251/5

= (xz + 1/5x)2 = 9

= (x + 1/5x) = √9

= ± 3

If (x + 1/5x) = 3

∴ x3 + 1/125x3

= 25 1/5

Hence, the value is 25 1/5

Question no – (5)

Solution :

(i) 102 ×  98

= (100 + 2) (100 – 2)

= (100)2 – (2)2

= 10000 – 4

= 9996

(ii) 10032 – 9972

= (1003 + 997) (1003 – 997)

= 2000 × 6

= 12000

(iii) (10)3 – (5)3 – (5)3

= 1000 – 125 – 125

= 750

Next Chapter Solution :

Updated: June 19, 2023 — 2:26 pm