ML Aggarwal Factorisation Exe-4.5 Class 9 ICSE Maths Solutions . Step by step solutions of Factorisation problems as council prescribe guideline. Visit official website CISCE for detail information about ICSE Board Class-9.
ML Aggarwal Factorisation Exe-4.5 Class 9 ICSE Maths Solutions
| Board | ICSE |
| Subject | Maths |
| Class | 9th |
| Chapter-4 | Factorisation |
| Topics | Solution of Exe-4.5 Questions |
| Academic Session | 2024-2025 |
Solution of Exe-4.5 Questions
ML Aggarwal Factorisation Exe-4.5 Class 9 ICSE Maths Solutions
Que-1:
Sol: (i) 8x3 + y3
Above terms can be written as,
(2x)3 + y3
We know that, a3 + b3 = (a + b) (a2 – ab + b2)
Where, a = 2x, b = y
Then, (2x)3 + y3 = (2x + y) ((2x)2 – (2x × y) + y2)
= (2x + y) (4x2 – 2xy + y2)
(ii) 64x3 – 125y3
Above terms can be written as,
(4x)3 – (5y)3
We know that, a3 – b3 = (a – b) (a2 + ab + b2)
Where, a = 4x, b = 5y
Then, (4x)3 – (5y)3 = (4x – 5y) ((4x)2 + (4x × 5y) + 5y2)
= (4x – 5y) (16x2 + 2oxy + 25y2)
Que-2:
Sol: (i) 64x3 + 1
Above terms can be written as,
(4x)3 + 13
We know that, a3 + b3 = (a + b) (a2 – ab + b2)
Where, a = 4x, b = 1
Then, (4x)3 + 13 = (4x + 1) ((4x)2 – (4x × 1) + 12)
= (4x + 1) (16x2 – 4x + 1)
(ii) 7a3 + 56b3
Take out common in all terms we get,
7(a3 + 8b3)
Above terms can be written as,
7(a3 + (2b)3)
We know that, a3 + b3 = (a + b) (a2 – ab + b2)
Where, a = a, b = 2b
Then, 7[(a)3 + (2b)3] = 7[(a + 2b) ((a)2 – (a × 2b) + (2b)2)]
= 7(a + 2b) (a2 – 2ab + 4b2)
Que-3:
Sol: (i) (x6/343) + (343/x6)
Above terms can be written as,
(x2/7)3 + (7/x2)3
We know that, a3 + b3 = (a + b) (a2 – ab + b2)
Where, a = (x2/7), b = (7/x2)
Then, (x2/7)3 + (7/x2)3 = [(x2/7) + (7/x2)] [(x2/7)2 – ((x2/7) × (7/x2)) + (7/x2)2]
= [(x2/7) + (7/x2)] [(x4/49) – 1 + (49/x4)]
(ii)8x3 – 1/27y3
Above terms can be written as,
(2x)3 – (1/3y)3
We know that, a3 – b3 = (a – b) (a2 + ab + b2)
Where, a = 2x, b = (1/3y)
Then, (2x)3 – (1/3y)3 = (2x – (1/3y)) ((2x)2 + (2x × (1/3y)) + (3y)2)
= (2x – (1/3y)) (4x2 + (2x/3y) + 9y2)
Que-4:
Sol: (i) x2 + x5
Take out common in all terms we get,
x2(1 + x3)
x2(13 + x3)
We know that, a3 + b3 = (a + b) (a2 – ab + b2)
Where, a = 1, b = x
= x2 [(1 + x) (12 – (1 × x) + x2)]
= x2 (1 + x) (1 – x + x2)
(ii)32x4 – 500x
Take out common in all terms we get,
4x(8x3 – 125)
Above terms can be written as,
4x((2x)3 – 53)
We know that, a3 – b3 = (a – b) (a2 + ab + b2)
Where, a = 2x, b = 5
= 4x(2x – 5) ((2x)2 + (2x × 5) + 52)
= 4x(2x – 5) (4x2 + 10x + 25)
Que-5:
Sol: (i) 27x3y3 – 8
Above terms can be written as,
(3xy)3 – 23
We know that, a3 – b3 = (a – b) (a2 + ab + b2)
Where, a = 3xy, b = 2
= (3xy – 2) ((3xy)2 + (3xy × 2) + 22)
= (3xy – 2) (9x2y2 + 6xy + 4)
(ii) 27(x + y)3 + 8(2x – y)3
Above terms can be written as,
33(x + y)3 + 23(2x – y)3
(3(x + y))3 + (2(x – y))3
We know that, a3 + b3 = (a + b) (a2 – ab + b2)
Where, a = 3(x + y), b = 2(x – y)
= [3(x + y) + 2(2x – y)] [(3(x + y))3 – (3(x + y) × 2(2x – y)) + (2(2x – y))2]
= [3x + 3y + 4x – 2y] [9(x + y)2 – 6(x + y)(2x – y) + 4(2x – y)2]
= (7x – y) [9(x2 + y2 + 2xy) – 6(2x2 – xy + 2xy – y2) + 4(4x2 + y2 – 4xy)]
= (7x – y) [9x2 + 9y2 + 18xy – 12x2 – 6xy – 6y2 + 16x2 + 4y2 – 16xy]
= (7x – y) [13x2 – 4xy + 19y2]
Que-6:
Sol: (i) a3 + b3 + a + b
We know that, a3 + b3 = (a + b) (a2 – ab + b2)
[(a + b) (a2 – ab + b2)] + (a + b)
(a + b) (a2 – ab + b2 + 1)
(ii) a3 – b3 – a + b
(a3 – b3) – (a – b)
We know that, a3 – b3 = (a – b) (a2 + ab + b2)
[(a – b) (a2 + ab + b2)] – (a – b)
(a – b) (a2 + ab + b2 – 1)
Que-7:
Sol: (i) x3 + x + 2
Above terms can be written as,
x3 + x + 1 + 1
Rearranging the above terms, we get
(x3 + 1) (x + 1)
(x3 + 13) (x + 1)
We know that, a3 + b3 = (a + b) (a2 – ab + b2)
[(x + 1) (x2 – x + 1)] + (x + 1)
(x + 1) (x2 – x + 1 + 1)
(x + 1) (x2 – x + 2)
(ii) a3 – a – 120
Above terms can be written as,
a3 – a – 125 + 5
Rearranging the above terms, we get
a3 – 125 – a + 5
(a3 – 125) – (a – 5)
(a3 – 53) – (a – 5)
We know that, a3 – b3 = (a – b) (a2 + ab + b2)
[(a – 5) (a2 + 5a + 52)] – (a – 5)
(a – 5) (a2 + 5a + 25) – (a – 5)
(a – 5) (a2 + 5a + 25 – 1)
(a – 5) (a2 + 5a + 24)
Que-8:
Sol: (i) x3 + 6x2 + 12x + 16
x3 + 6x2 + 12x + 8 + 8
Above terms can be written as,
(x3 + (3 × 2 × x2) + (3 × 22 × x) + 23) + 8
We know that, (a + b)3 = a3 + b3 + 3a2b + 3ab2
Now a = x and b = 2
So, (x + 2)3 + 23
We know that, a3 + b3 = (a + b) (a2 – ab + b2)
(x + 2 + 2) ((x + 2)2 – (2 × (x + 2)) + 22)
(x + 4) (x2 + 4 + 4x – 2x – 4 + 4)
(x + 4) (x2 + 2x + 4)
(ii) a3 – 3a2b + 3ab2 – 2b3
Above terms can be written as,
a3 – 3a2b + 3ab2 – b3 – b3
We know that, (a – b)3 = a3 – b3 – 3a2b + 3ab2
So, (a – b)3 + b3
We also know that, a3 – b3 = (a – b) (a2 + ab + b2)
Where, a = a – b, b = b
(a – b – b) ((a – b)2 + (a – b)b + b2)
(a – 2b) (a2 + b2 – 2ab + ab – b2 + b2)
(a – 2b) (a2 + b2 – ab)
Que-9:
Sol: (i) 2a3 + 16b3 – 5a – 10b
Above terms can be written as,
2(a3 + 8b3) – 5(a + 2b)
2(a3 + (2b)3) – 5(a + 2b)
We know that, a3 + b3 = (a + b) (a2 – ab + b2)
2[(a + 2b) (a2 – 2ab + 4b2)] – 5(a + 2b)
(a + 2b) (2a2 – 4ab + 8b2 – 5)
(ii) a3 – (1/a3) – 2a + 2/a
(a3 – (1/a)3) – 2a + 2/a
We know that, a3 – b3 = (a – b) (a2 + ab + b2)
[(a – 1/a) – (a2 + (a × 1/a) + (1/a)2] – 2(a – 1/a)
(a – 1/a) (a2 + 1 + 1/a2) – 2(a – 1/a)
(a – 1/a) (a2 + 1 + 1/a2 – 2)
(a – 1/a) (a2 + (1/a2) – 1)
Que-10:
Sol: (i) a6 – b6
Above terms can be written as,
(a2)3 – (b2)3
We know that, a3 – b3 = (a – b) (a2 + ab + b2)
So, a = a2, b = b2
(a2 – b2) ((a2)2) + a2b2 + (b2)2)
(a2 – b2) (a4 + a2b2 + b4)
(ii) x6 – 1
Above terms can be written as,
(x2)3 – 13
We know that, a3 – b3 = (a – b) (a2 + ab + b2)
So, a = x2, b = 1
(x2 – 1) ((x2)2 + (x2 × 1) + 12)
(x2 – 1) (x4 + x2 + 1)
Que-11:
Sol: (i) 64x6 – 729y6
Above terms can be written as,
(2x)6 – (3y)6
[(2x)2]3 – [(3y)2]3
We know that, a3 – b3 = (a – b) (a2 + ab + b2)
So, a = (2x)2, b = (3y)2
[(2x)2 – (3y)2] [((2x)2)2 + ((2x)2× (3y)2) + ((3y)2)2]
(4x2 – 9y2) [16x4 + (4x2 × 9y2) + (9y2)2]
(4x2 – 9y2) [16x4 + 36x2y2 + 81y4] [(2x)2 – (3y)2] [16x4 + 36x2y2 + 81y4]
(2x + 3y) (2x – 3y) (16x4 + 36x2y2 + 81y4)
(ii) x3 – (8/x)
Above terms can be written as,
(1/x) (x3 – 8)
(1/x) [(x)3 – (2)3]
We know that, a3 – b3 = (a – b) (a2 + ab + b2)
So, a = x, b = 2
(1/x) (x – 2) (x2 + 2x + 4)
Que-12:
Sol: (i) 250 (a – b)3 + 2
Take out common in all terms we get,
2(125(a – b)3 + 1)
2[(5(a – b))3 + 13]
We know that, a3 + b3 = (a + b) (a2 – ab + b2)
= 2[(5a – 5b + 1) ((5a – 5b)2 – (5a – 5b)1 + 12)]
= 2(5a – 5b + 1) (25a2 + 25b2 – 50ab – 5a + 5b + 1)
(ii) 32a2x3 – 8b2x3 – 4a2y3 + b2y3
Take out common in all terms we get,
8x3(4a2 – b2) – y3(4a2 – b2)
(4a2 – b2) (8x3 – y3)
Above terms can be written as,
((2a)2 – b2) ((2x)3 – y3)
We know that, a3 – b3 = (a – b) (a2 + ab + b2) and (a2 – b2) = (a + b) (a – b)
(2a + b) (2a – b) [(2x – y) ((2x)2 + 2xy + y2)]
(2a + b) (2a – b) (2x – y) (4x2 + 2xy + y2)
Que-13:
Sol: (i) x9 + y9
Above terms can be written as,
(x3)3 + (y3)3
We know that, a3 + b3 = (a + b) (a2 – ab + b2)
Where, a = x3, b = y3
(x3 + y3) ((x3)2 – x3y3 + (y3)2)
(x3 + y3) (x6 – x3y3 + y6)
Then, (x3 + y3) in the form of (a3 + b3)
(x + y)(x2 – xy + y2) (x6 – x3y3 + y6)
(ii) X6 – 7x3 – 8
Above terms can be written as,
(x2)3 – 7x3 – x3 + x3 – 8
(x2)3 – 8x3 + x3 – 23
(((x2)3) – (2x)3) + (x3 – 23)
We know that, a3 – b3 = (a – b) (a2 + ab + b2)
(x2 – 2x) ((x2)2 + (x2 × 2x) + (2x)2) + (x – 2) (x2 + 2x + 22)
(x2 – 2x) (x4 + 2x3 + 4x2) + (x – 2) (x2 + 2x + 4)
x(x – 2) x2(x2 + 2x + 4) + (x – 2) (x2 + 2x + 4)
Take out common in all terms we get,
(x – 2) (x2 + 2x + 4) ((x × x2) + 1)
(x – 2) (x2 + 2x + 4) (x3 + 1)
So, above terms are in the form of a3 + b3
Therefore, (x – 2) (x2 + 2x + 4) (x + 1) (x2 – x + 1)
— : End of ML Aggarwal Factorisation Exe-4.5 Class 9 ICSE Maths Solutions :–
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