ML Aggarwal Expansions Chapter Test Class 9 ICSE Maths Solutions

ML Aggarwal Expansions Chapter Test Class 9 ICSE Maths Solutions. We Provide Step by Step Answer of  Chapter Test Questions for Expansions as council prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ICSE Board Class-9.

ML Aggarwal Expansions Chapter Test Class 9 ICSE Maths Solutions

Board ICSE
Subject Maths
Class 9th
Chapter-3 Expansions
Topics Solution of Ch-Test Questions
Edition 2024-2025

Solution of Ch-Test Questions

ML Aggarwal Expansions Chapter Test Class 9 ICSE Maths Solutions

Question 1. Find the expansions of the following :

(i) (2x + 3y + 5) (2x + 3y – 5)
(ii) (6 – 4a -7b)2
(iii) (7 – 3xy)3
(iv) (x + y + 2)3

Answer :

(i) (2x + 3y + 5) (2x + 3y – 5)

Let us simplify the expression, we get

(2x + 3y + 5) (2x + 3y – 5) = [(2x + 3y) + 5] [(2x – 3y) – 5]

By using the formula, (a)2 – (b)2 = [(a + b) (a – b)]
= (2x + 3y)2 – (5)2

= (2x)2 + (3y) 2 + 2 × 2x × 3y – 5 × 5

= 4x2 + 9y2 + 12xy – 25

(ii) (6 – 4a – 7 b)2

Let us simplify the expression, we get

(6 – 4a – 7 b)2 = [ 6 + (- 4a) + (-7b)]2

= (6)2 + (- 4a)2 + (- 7b)2 + 2 (6) (- 4a) + 2 (- 4a) (-7b) + 2 (-7b) (6)

= 36 + 16a2 + 49b2 – 48a + 56ab – 84b

(iii) (7 – 3xy)3

Let us simplify the expression

By using the formula, we get

(7 – 3xy)= (7)3 – (3xy)3 – 3 (7) (3xy) (7 – 3xy)

= 343 – 27x3y3 – 63xy (7 – 3xy)

= 343 – 27x3y3 – 441xy + 189x2y2

(iv) (x + y + 2)3

Let us simplify the expression

By using the formula, we get

(x + y + 2 )3 = [(x + y) + 2]3

= (x + y)3 + (2)3 + 3 (x + y) (2) (x + y + 2)

= x3 + y3 + 3x2y + 3xy2 + 8 + 6 (x + y) [(x + y) + 2]
= x3 + y3 + 3x2y + 3xy2 + 8 + 6 (x + y)2 +12(x + y)

= x3 + y3 + 3x2y + 3xy2 + 8 + 6 (x2 + y2 + 2xy) + 12x + 12y = x3 + y3 + 3x2y + 3xy2 + 8 + 6x2 + 6y2 + 12xy + 12x + 12y

= x3 + y3 + 3x2y + 3xy2 + 8 + 6x2 + 6y2 + 12x + 12y + 12xy

Question 2. Simplify: (x – 2) (x + 2) (x2 + 4) (x4 + 16)

Answer :

Let us simplify the expression, we get

(x – 2) (x + 2) (x4 + 4) (x4 +16) = (x2 – 4) (x4 + 4) (x4 + 16)

= [(x2)2 – (4)2] (x4 + 16)

= (x4 – 16) (x4 + 16)

= (x4)2 – (16)2

= x8 – 256

Question 3. Evaluate 1002 × 998 by using a special product.
Answer :

Let us simplify the expression, we get

1002 × 998 = (1000 + 2) (1000 – 2)

= (1000)2 – (2)2

= 1000000 – 4

= 999996

Question 4. If a + 2b + 3c = 0, Prove that a3 + 8b3 + 27c3 = 18 abc
Answer :

Given:

a + 2b + 3c = 0, a + 2b = – 3c

Let us cube on both the sides, we get

(a + 2b)3 = (-3c)3

a3 + (2b)3 + 3(a) (2b) (a + 2b) = -27c3

a3 + 8b3 + 6ab (– 3c) = – 27c3

a3 + 8b3 – 18abc = -27c3

a3 + 8b3 + 27c3 = 18abc

Hence proved.

Question 5. If 2x = 3y – 5, then find the value of 8x3 – 27y3 + 90xy + 125.
Answer :

Given:

2x = 3y – 5

2x – 3y = -5

Now, let us cube on both sides, we get

(2x – 3y)3 = (-5)3

(2x)3 – (3y)3 – 3 × 2x × 3y (2x – 3y) = -125

8x3 – 27y3 – 18xy (2x – 3y) = -125

Now, substitute the value of 2x – 3y = -5

8x3 – 27y3 – 18xy (-5) = -125

8x3 – 27y3 + 90xy = -125

8x3 – 27y3 + 90xy + 125 = 0

Question 6. If a2 – 1/a2 = 5, evaluate a4 + 1/a4

Answer :

It is given that,

a2 – 1/a2 = 5

So,

By using the formula, (a + b)2

[a2 – 1/a2]2 = a4 + 1/a4 – 2

[a2 – 1/a2]2 + 2 = a4 + 1/a4

Substitute the value of a2 – 1/a2 = 5, we get

52 + 2 = a4 + 1/a4

a4 + 1/a4 = 25 + 2

= 27

Question 7. If a + 1/a = p and a – 1/a = q, Find the relation between p and q.

Answer :

It is given that,

a + 1/a = p and a – 1/a = q

so,

(a + 1/a)2 – (a – 1/a)2 = 4(a) (1/a)

= 4

By substituting the values, we get

p2 – q2 = 4

Hence the relation between p and q is that p2 – q2 = 4.

Question 8. If (a2 + 1)/a = 4, find the value of 2a3 + 2/a3

Answer :

It is given that,

(a2 + 1)/a = 4

a2/a + 1/a = 4

a + 1/a = 4

So by multiplying the expression by 2a, we get

2a3 + 2/a3 = 2[a3 + 1/a3]

= 2 [(a + 1/a)3 – 3 (a) (1/a) (a + 1/a)]

= 2 [(4)3 – 3(4)]

= 2 [64 – 12]

= 2 (52)

= 104

Question 9. If x = 1/(4 – x), find the value of

(i) x + 1/x

(ii) x3 + 1/x3

(iii) x6 + 1/x6

Answer 9

It is given that,

x = 1/(4 – x)

So,

(i) x(4 – x) = 1

4x – x2 = 1

Now let us divide both sides by x, we get

4 – x = 1/x

4 = 1/x + x

1/x + x = 4

1/x + x = 4

(ii) x3 + 1/x3 = (x + 1/x)2 – 3(x + 1/x)

By substituting the values, we get

= (4)3 – 3(4)

= 64 – 12

= 52

(iii) x6 + 1/x6 = (x3 + 1/x3)2 – 2

= (52)2 – 2

= 2704 – 2

= 2702

Question 10. If x – 1/x = 3 + 2√2, find the value of ¼ (x3 – 1/x3)

Answer :

It is given that,

x – 1/x = 3 + 2√2

So,

x3 – 1/x3 = (x – 1/x)3 + 3(x – 1/x)

= (3 + 2√2)3 + 3(3 + 2√2)

By using the formula, (a+b)3 = a3 + b3 + 3ab (a + b)

= (3)3 + (2√2)3 + 3 (3) (2√2) (3 + 2√2) + 3(3 + 2√2)

= 27 + 16√2 + 54√2 + 72 + 9 + 6√2

= 108 + 76√2

Hence,

¼ (x3 – 1/x3) = ¼ (108 + 76√2)

= 27 + 19√2

Question 11. If x + 1/x = 3 1/3, find the value of x3 – 1/x3

Answer :

It is given that,

x + 1/x = 3 1/3

we know that,

(x – 1/x)2 = x2 + 1/x2 – 2

= x2 + 1/x2 + 2 – 4

= (x + 1/x)2 – 4

But x + 1/x = 3 1/3 = 10/3

So,

(x – 1/x)2 = (10/3)2 – 4

= 100/9 – 4

= (100 – 36)/9

= 64/9

x – 1/x = √(64/9)

= 8/3

Now,

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

= (8/3)3 + 3 (8/3)

= ((512/27) + 8)

= 728/27

= 26 26/27

Question 12. If x = 2 – √3, then find the value of x3 – 1/x3

Answer :

It is given that,

x = 2 – √3

so,

1/x = 1/(2 – √3)

By rationalizing the denominator, we get

= [1(2 + √3)] / [(2 – √3) (2 + √3)]

= [(2 + √3)] / [(22) – (√3)2]

= [(2 + √3)] / [4 – 3]

= 2 + √3

Now,

x – 1/x = 2 – √3 – 2 – √3

= – 2√3

Let us cube on both sides, we get

(x – 1/x)3 = (-2√3)3

x3 – 1/x3 – 3 (x) (1/x) (x – 1/x) = 24√3

x3 – 1/x3 – 3(-2√3) = -24√3

x3 – 1/x3 + 6√3 = -24√3

x3 – 1/x3 = -24√3 – 6√3

= -30√3

Hence,

x3 – 1/x3 = -30√3

Question 13. If the sum of two numbers is 11 and sum of their cubes is 737, find the sum of their squares.
Answer :

Let us consider x and y be two numbers

Then,

x + y = 11

x3 + y3 = 735 and x2 + y2 =?

Now,

x + y = 11

Let us cube on both the sides,

(x + y)3 = (11)3

x3 + y3 + 3xy (x + y) = 1331

737 + 3x × 11 = 1331

33xy = 1331 – 737

= 594

xy = 594/33

xy = 8

We know that, x + y = 11

By squaring on both sides, we get

(x + y)2 = (11)2

x2 + y2 + 2xy = 121 2 x2 + y2 + 2 × 18 = 121

x2 + y2 + 36 = 121

x2 + y2 = 121 – 36

= 85

Hence sum of the squares = 85

Question 14. If a – b = 7 and a3 – b3 = 133, find:
(i) ab
(ii) a2 + b2
Answer :

It is given that,

a – b = 7

let us cube on both sides, we get

(i) (a – b)3 = (7)3

a3 + b3 – 3ab (a – b) = 343

133 – 3ab × 7 = 343

133 – 21ab = 343

– 21ab = 343 – 133 21ab

= 210

ab = -210/21

ab = -10

(ii) a2 + b2

Again a – b = 7

Let us square on both sides, we get

(a – b)2 = (7)2

a2 + b2 – 2ab = 49

a2 + b2 – 2 × (- 10) = 49

a2 + b2 + 20 = 49

a2 + b2 = 49 – 20

= 29

Hence, a2 + b2 = 29

Question 15. Find the coefficient of x2 expansion of (x2 + x + 1)2 + (x2 – x + 1)2
Answer :

Given:

The expression, (x2 + x + 1)2 + (x2 – x + 1)2

(x2 + x + 1)2 + (x2 – x + 1)2 = [((x2 + 1) + x)2 + [(x2 + 1) – x)2]
= (x2 + 1)2 + x2 + 2 (x2 + 1) (x) + (x2 + 1)2 + x2 – 2 (x2 + 1) (x)

= (x2)2 + (1)2 + 2 × x2 × 1 + x2 + (x2)2 + 1 + 2 × x2 + 1 + x2

= x4 + 1 + 2x2 + x2 + x4 + 1 + 2x2 + x2

= 2x4 + 6x2 + 2

∴ Co-efficient of x2 is 6.

—  : End of ML Aggarwal Expansions Chapter Test Class 9 ICSE Maths Solutions :–

Return to :-  ML Aggarawal Maths Solutions for ICSE  Class-9

Thanks

Please Share with Your Friends

Leave a Comment

This site uses Akismet to reduce spam. Learn how your comment data is processed.