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)3 = (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 :–
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