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)