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Chapter 5: Factorization of Algebraic Expressions Exercise – 5.3 Question: 1 Find the value 64a3 + 125b3 + 240a2b + 300ab2 Solution: = (4a)3 + (5b)3 + 3(4a)2(5b) + 3(4a)(5b)2 [∴ a3 + b3 + 3a2b + 3ab2 = (a + b)3] = (4a + 5b)3 = (4a + 5b)(4a + 5b)(4a + 5b) ∴ 64a3 + 125b3 + 240a2b + 300ab2 = (4a + 5b)(4a + 5b)(4a + 5b) Question: 2 Find the value 125x3 − 27y3 − 225x2y + 135xy2 Solution: = (5x)3− (3y)3 − 3(5x)2(3y) + 3(5x)(3y)2 [∴ a3 − b3 − 3a2b + 3ab2 = (a − b)3] = (5x − 3y)3 = (5x − 3y)(5x − 3y)(5x − 3y) ∴ 125x3 − 27y3 − 225x2y + 135xy2 = (5x − 3y)(5x − 3y)(5x − 3y) Question: 3 Find the value Solution: [∴ x3 + b3 + 3x2b + 3xb2 = (x + b)3] Question: 4 Find the value 8x3 + 27y3 + 36x2y + 54xy2 Solution: = (2x)3 + (3y)3 + 3 × (2x)2 × 3y + 3 × (2x)(3y)2 = (2x + 3y)3 [∴ a³ + b³ + 3a²b + 3ab² = (a + b)³] = (2x + 3y)(2x + 3y)(2x + 3y) ∴ 8x3 + 27y3 + 36x2y + 54xy2 = (2x + 3y)(2x + 3y)(2x + 3y) Question: 5 Find the value a3 − 3a2b + 3ab2 − b3 + 8 Solution: = (a − b)3 + 23 [∴ a3 − b3 − 3a2b + 3ab2 = (a − b)3] = (a − b + 2)((a − b)2 − (a − b)2 + 22) ∴ [a3 + b3 = (a + b)(a2 − ab + b2)] = (a − b + 2)(a2 + b2 − 2ab − 2(a − b) + 4) = (a − b + 2)(a2 + b2 − 2ab − 2a + 2b + 4) ∴ a3 − 3a2b + 3ab2 − b3 + 8 = (a − b + 2)(a2 + b2 − 2ab − 2a + 2b + 4) Question: 6 Find the value x3 + 8y3 + 6x2y + 12xy2 Solution: = (x)3 + (2y)3 + 3 × x2 × 2y + 3 × x × (2y)2 = (x + 2y)3 [∴ x3 + y3 + 3x2y+3xy2 = (x + y)3] = (x + 2y)(x + 2y)(x + 2y) ∴ x3 + 8y3 + 6x2y + 12xy2 = (x + 2y)(x + 2y)(x + 2y) Question: 7 Find the value 8x3 + y3 + 12x2y + 6xy2 Solution: = (2x)3 + (y)3 + 3 × (2x)2 × y + 3(2x) × y2 = (2x + y)3 [∴ a3 + b3 + 3a2b + 3ab2 = (a + b)3] = (2x + y)(2x + y)(2x + y) ∴ 8x3 + y3 + 12x2y + 6xy2 = (2x + y)(2x + y)(2x + y) Question: 8 Find the value 8a3 + 27b3 + 36a2b + 54ab2 Solution: = (2a)3 + (3b)3 + 3 × (2a)2 × 3b + 3 × 2a × (3b)2 = (2a + 3b)3 [∴ a3 + b3 + 3a2b + 3ab2 = (a + b)3] = (2a + 3b)(2a + 3b)(2a + 3b) ∴ 8a3 + 27b3 + 36a2b + 54ab2 = (2a + 3b)(2a + 3b)(2a + 3b) Question: 9 Find the value 8a3 − 27b3 − 36a2b + 54ab2 Solution: = (2a)3 − (3b)3 − 3 × (2a)2 × 3b + 3 × 2a × (3b)2 = (2a − 3b)3 [∴ a3 − b3 − 3a2b + 3ab2 = (a − b)3] = (2a − 3b)(2a − 3b)(2a − 3b) ∴ 8a3 − 27b3 − 36a2b + 54ab2 = (2a − 3b)(2a − 3b)(2a − 3b) Question: 10 Find the value x3 − 12x(x − 4) − 64 Solution: = x3 − 12x2 + 48x − 64 = x3 − 3 × x2 × 4 + 3 × 42 × x − 43 = (x − 4)3 [∴ a3 − b3 − 3a2b + 3ab2 = (a − b)3] = (x − 4)(x −4 )(x − 4) ∴ x3 − 12x(x − 4) − 64 = (x − 4)(x − 4)(x − 4) Question: 11 Find the value a3x3 − 3a2bx2 + 3ab2x − b3 Solution: = (ax)3 − 3(ax)2 × b + 3(ax) × b2 − b3 = (ax − b)3 [∴ a3 − b3 − 3a2b + 3ab2 = (a − b)3] = (ax − b)(ax − b)(ax − b) ∴ a3x3 − 3a2bx2 + 3ab2x − b3 = (ax − b)(ax − b)(ax − b)
Find the value
64a3 + 125b3 + 240a2b + 300ab2
= (4a)3 + (5b)3 + 3(4a)2(5b) + 3(4a)(5b)2
[∴ a3 + b3 + 3a2b + 3ab2 = (a + b)3]
= (4a + 5b)3
= (4a + 5b)(4a + 5b)(4a + 5b)
∴ 64a3 + 125b3 + 240a2b + 300ab2
125x3 − 27y3 − 225x2y + 135xy2
= (5x)3− (3y)3 − 3(5x)2(3y) + 3(5x)(3y)2
[∴ a3 − b3 − 3a2b + 3ab2 = (a − b)3]
= (5x − 3y)3
= (5x − 3y)(5x − 3y)(5x − 3y)
∴ 125x3 − 27y3 − 225x2y + 135xy2
[∴ x3 + b3 + 3x2b + 3xb2 = (x + b)3]
8x3 + 27y3 + 36x2y + 54xy2
= (2x)3 + (3y)3 + 3 × (2x)2 × 3y + 3 × (2x)(3y)2
= (2x + 3y)3
[∴ a³ + b³ + 3a²b + 3ab² = (a + b)³]
= (2x + 3y)(2x + 3y)(2x + 3y)
∴ 8x3 + 27y3 + 36x2y + 54xy2
a3 − 3a2b + 3ab2 − b3 + 8
= (a − b)3 + 23
= (a − b + 2)((a − b)2 − (a − b)2 + 22)
∴ [a3 + b3 = (a + b)(a2 − ab + b2)]
= (a − b + 2)(a2 + b2 − 2ab − 2(a − b) + 4)
= (a − b + 2)(a2 + b2 − 2ab − 2a + 2b + 4)
∴ a3 − 3a2b + 3ab2 − b3 + 8
x3 + 8y3 + 6x2y + 12xy2
= (x)3 + (2y)3 + 3 × x2 × 2y + 3 × x × (2y)2
= (x + 2y)3
[∴ x3 + y3 + 3x2y+3xy2 = (x + y)3]
= (x + 2y)(x + 2y)(x + 2y)
∴ x3 + 8y3 + 6x2y + 12xy2 = (x + 2y)(x + 2y)(x + 2y)
8x3 + y3 + 12x2y + 6xy2
= (2x)3 + (y)3 + 3 × (2x)2 × y + 3(2x) × y2
= (2x + y)3
= (2x + y)(2x + y)(2x + y)
∴ 8x3 + y3 + 12x2y + 6xy2 = (2x + y)(2x + y)(2x + y)
8a3 + 27b3 + 36a2b + 54ab2
= (2a)3 + (3b)3 + 3 × (2a)2 × 3b + 3 × 2a × (3b)2
= (2a + 3b)3
= (2a + 3b)(2a + 3b)(2a + 3b)
∴ 8a3 + 27b3 + 36a2b + 54ab2
8a3 − 27b3 − 36a2b + 54ab2
= (2a)3 − (3b)3 − 3 × (2a)2 × 3b + 3 × 2a × (3b)2
= (2a − 3b)3
= (2a − 3b)(2a − 3b)(2a − 3b)
∴ 8a3 − 27b3 − 36a2b + 54ab2
x3 − 12x(x − 4) − 64
= x3 − 12x2 + 48x − 64
= x3 − 3 × x2 × 4 + 3 × 42 × x − 43
= (x − 4)3
= (x − 4)(x −4 )(x − 4)
∴ x3 − 12x(x − 4) − 64 = (x − 4)(x − 4)(x − 4)
a3x3 − 3a2bx2 + 3ab2x − b3
= (ax)3 − 3(ax)2 × b + 3(ax) × b2 − b3
= (ax − b)3
= (ax − b)(ax − b)(ax − b)
∴ a3x3 − 3a2bx2 + 3ab2x − b3
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Chapter 5: Factorization of Algebraic Expressions...