Factorisation Class 9 RS Aggarwal Exe-4C Goyal Brothers ICSE Maths Solutions Ch-4. In this article you will learn how to Factorise Sum or Differences of Cubes easily using formula. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics.
Factorisation Class 9 RS Aggarwal Exe-4C Goyal Brothers ICSE Maths Solutions Ch-4
| Board | ICSE |
| Subject | Maths |
| Class | 9th |
| Chapter-4 | Factorisation |
| Topics | Factorise Sum or Differences of Cubes using formula |
| Academic Session | 2024-2025 |
Factorise Sum or Differences of Cubes using formula
The sum or difference of two cubes can be factored into a product of a binomial times a trinomial. That is, x 3 + y 3 = ( x + y ) ( x 2 − x y + y 2 ) and x 3 − y 3 = ( x − y ) ( x 2 + x y + y 2 )
Exercise- 4C
Factorise : Page- 63,64
Que-1: x³+64
Sol: x³+64
(x)³+(4)³
(x+4)[(x)²-x×4+(4)²]
(x+4)(x²-4x+16).
Que-2: 8a³+27b³
Sol: 8a³+27b³
(2a)³+(3b)³
(2a+3b)[(2a)²-2a x 3b+(3b)²]
(2a+3b)(4a²-6ab+9b²)
Que-3: 7a³+56b³
Sol: 7a³+56b³
7(a³+8b³)
7(a)³+(2b)³
7[(a+2b){(a)²-a x 2b+(2b)²}]
7(a+2b)(a²-2ab+4b²)
Que-4: x^5 + x²
Sol: x^5 + x²
x²(x³+1)
x²[(x)³+(1)³]
x²[(x+1){(x)²-x × 1 +1²}]
x²(x+1)(x²-x+1).
Que-5: 16x^4 + 54x
Sol: 16x^4 + 54x
2x(8x³+27)
2x[(2x)³+(3)³]
2x[(2x+3){(2x)²-2x x 3 +(3)²}]
2x(2x+3)(4x²-6x+9).
Que-6: 216x³ + (1/27)
Sol: 216x³+(1/27)
(6x)³+(1/3)³
(6x+(1/3))[(6x)²-6x x 1/3 + (1/3)²]
(6x+1/3)(36x²-2x+1/9).
Que-7: a^6 + b^6
Sol: a^6 + b^6
(a²)³+(b²)³
(a²+b²)[(a²)²-a²xb²+(b²)²]
(a²+b²)(a^4 – a²b² + b^4).
Que-8: a^4 + 343a
Sol: a^4 + 343a
a(a³+343)
a[(a)³+(7)³]
a(a+7)[(a)²-ax7+(7)²]
a(a+7)(a²-7a+49).
Que-9: 125x³+1
Sol: 125x³+1
(5x)³+(1)³
(5x+1)[(5x)²-5x x 1 +(1)²]
(5x+1)(25x²-5x+1).
Que-10: 2a³+16b³-3a-6b
Sol: 2a³+16b³-3a-6b
2(a³+8b³)-3(a+2b)
2[(a)³+(2b)³]-3(a+2b)
2(a+2b)[(a)²-ax2b+(2b)²]-3(a+b)
2(a+2b)(a²-2ab+4b²-3)
(a+2b)[2(a²-2ab+4b²)-3].
Que-11: a³-125-2a+10
Sol: a³-125-2a+10
(a)³-(5)³-2(a-5)
(a-5)[(a)²+5xa+(5)²]-2(a-5)
(a-5)(a²+5a+25-2)
(a-5)(a²+5a+23).
Que-12: x³-125
Sol: x³-125
(x)³-(5)³
(x-5)[(x)²+5x+(5)²]
(x-5)(x²+5x+25).
Que-13: 8a³-(1/27b³)
Sol: 8a³-(1/27b³)
(2a)³-(1/3b)³
(2a-1/3b)[(2a)²+(2a/3b)+(1/3b)²]
(2a-1/3b)[4a²+(2a/3b)+(1/9b²)].
Que-14: (8a³/27) – (b³/8)
Sol: (8a³/27)-(b³/8)
(2a/3)³-(b/2)³
[(2a/3)-(b/2)][(2a/3)²+(ab/3)+(b/2)²]
[(2a/3)-(b/2)] [(4a²/9)+(ab/3)+(b²/4)].
Que-15: a-8ab³
Sol: a-8ab³
a(1-8b³)
a[(1)³-(2b)³]
a(1-2b)[(1)²+2b+(2b)²]
a(1-2b)(1+2b+4b²).
Que-16: x^6 – 1
Sol: x^6 – 1
(x³)²-(1)²
(x³+1)(x³-1)
[(x)³+(1)³] [(x)³-(1)³]
(x+1)[x²-x+1] (x-1)[(x)²+x+1]
(x+1)(x-1)(x²-x+1)(x²+x+1).
Que-17: a³-0.064
Sol: a³-0.064
(a)³-(0.4)³
(a-0.4)[(a)²+0.4a+(0.4)²]
(a-0.4)(a²+0.4a+0.16).
Que-18: 24x^4 – 375x
Sol: 24x^4 – 375x
3x(8x³-125)
3x[(2x)³-(5)³]
3x(2x-5)[(2x)²+10x+(5)²]
3x(2x-5)(4x²+10x+25).
Que-19: 3a^7 b – 81a^4 b^4
Sol: 3a^7 b – 81a^4 b^4
3a^4 b(a³-27b³)
3a^4 b[(a)³-(3b)³]
3a^4 b(a-3b)[(a)²+3a+(3b)²]
3a^4 b (a-3b)(a²+3a+9b²).
Que-20: a³-(1/a³)-2a+(2/a)
Sol: a³-(1/a)³-2a+(2/a)
[(a)³-(1/a)³]-2[a-(1/a)]
[a-(1/a)] [(a)²+1+(1/a)²]-2[a-(1/a)]
[a-(1/a)][a²+1+(1/a²)-2]
[a-(1/a)][a²+(1/a²)-1].
Que-21: 2x^7 – 128x
Sol: 2x(x^6 – 64)
2x[(x³)²-(8)²]
2x(x³+8)(x³-8)
2x[(x³)+(2)³][(x³)-(2)³]
2x(x+2)[(x)²-2x+(2)²](x-2)[(x)²+2x+(2)²]
2x(x+2)(x²-2x+4)(x-2)(x²+2x+4).
Que-22: 250(a-b)³ + 2
Sol: 250(a – b)³ + 2
= 2[ 125(a – b)³ + 1 ]
= 2 [ 5³(a – b)³ + 1 ]
= 2 [ (5a – 5b)³ + 1³ ]
= 2 (5a – 5b + 1) { (5a – 5b)² – (5a – 5b)×1 + 1² }
= 2 (5a – 5b + 1) (25a² – 50ab + 25b² – 5a + 5b + 1).
Que-23: 8a³-b³-4ax+2bx
Sol: 8a³-b³-4ax+2bx
We know that,
a³ – b³ = ( a – b )( a² + ab + b² ) …..(1)
8a³ – b³ – 4ax + 2bx
= [ (2a)³ – (b)³ ] – 2 x ( 2a – b )
= ( 2a – b )[ (2a)² + 2a x b + (b)² ] – 2 x ( 2a – b ) [ From(1) ]
= ( 2a – b )[ 4a² + 2ab + b² ] – 2 x ( 2a – b )
= ( 2a – b )[ 4a² + 2ab + b² – 2x ]
Que-24: a³-27b³+2a²b-6ab²
Sol: a³ – 27b³ + 2a²b – 6ab²
We know that,
a³ – b³ = ( a – b )( a² + ab + b² ) ….(1)
a³ – 27b³ + 2a²b – 6ab²
= (a)³ – (3b)³ + 2ab( a – 3b )
= ( a – 3b )[ a² + a x 3b + (3b)² ] + 2ab( a – 3b ) [From(1)]
= ( a – 3b )[ a² + 3ab + 9b² ] + 2ab( a – 3b )
= ( a – 3b )[ a² + 3ab + 9b² + 2ab ]
= ( a – 3b )[ a² + 5ab + 9b² ]
Que-25: 32a²x³-8b²x³-4a²y³+b²y³
Sol: 32𝑎²𝑥²−8𝑏²𝑥³−4𝑎²𝑦³+𝑏²𝑦³
Take out common in all terms we get,
8𝑥³(4𝑎²−𝑏²)−𝑦³(4𝑎²−𝑏²)
(4𝑎²−𝑏²)(8𝑥³−𝑦³)
Above terms can 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-26: a²-4b²+a³-8b³-(a-2b)²
Sol: a²−4b²+a³−8b³−(a−2b)²
= a²−4b²+a³−8b³−a²−4b²+4ab
= a³−8b²−8b³+4ab
= a³−(2b)³+4ab−8b²
= (a−2b)(a²+2ab+4b²)+4b(a−2b)
= (a−2b)(a²+2ab+4b²+4b)
Que-27: (a+b)³ + (a-b)³
Sol: (a+b)³ + (a-b)³
Let (a+b) = x and (a-b) = y
(a+b)³ + (a-b)³ = x³+y³
(x+y)(x²-xy+y²)
(a+b+a-b)[(a+b)²-(a+b)(a-b)+(a-b)²]
2a[(a+b)²+(a-b)²-(a²-b²)]
2a[2(a²+b²)-(a²-b²)]
2a[2a²+2b²-a²+b²]
2a(a²+3b²)
Que-28: x³-3x²+3x+7
Sol: 𝑥³-3𝑥²+3𝑥+7
𝑥³-3𝑥²+3𝑥+7
= 𝑥³+𝑥²-4𝑥²-4𝑥+7𝑥+7
= 𝑥²(𝑥+1)-4𝑥(𝑥+1)+7(𝑥+1)
= (𝑥+1)(𝑥²-4𝑥+7)
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