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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)

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