Chapter 5: Factorization of Algebraic Expressions Exercise – 5.2

Question: 1

Find the value

p³ + 27

Solution:

= p³ + 3³                                       

∴ [a³ + b³ = (a + b)(a² − ab + b²)]

= (p + 3)(p² - 3p - 9)

∴ p³ + 27 = (p + 3)(p² - 3p - 9)

Question: 2

Find the value

y³ + 125

Solution:

= y³ + 5³                                       

∴ [a³ + b³ = (a + b)(a² − ab + b²)]

= (y + 5)(y² − 5y + 5²)

= (y + 5)(y² − 5y + 25)

∴ y³ + 125 = (y + 5)(y² − 5y + 25)

Question: 3

Find the value

1 − 27a³

Solution:

= (1)³ − (3a)³

= (1 − 3a)(1² + 1 × 3a + (3a)²)                  

∴ [a³ − b³ = (a − b)(a² + ab + b²)]

= (1 − 3a)(1² + 3a + 9a²)

∴ 1 − 27a³ = (1 − 3a)(1² + 3a + 9a²)

Question: 4

Find the value

8x³y³ + 27a³

Solution:

= (2xy)³ + (3a)³

= (2xy + 3a)((2xy)² − 2xy × 3a + (3a)²) 

∴ [a³ + b³ = (a + b)(a² − ab + b²)]

= (2xy + 3a)(4x²y² − 6xya + 9a²)

∴ 8x³y³ + 27a³ = (2xy + 3a)(4x²y² − 6xya + 9a²)

Question: 5

Find the value

64a³ − b³

Solution:

= (4a)³ − b³

= (4a − b)((4a)² + 4a × b + b²)

∴ [a³ − b³ = (a − b)(a² + ab + b²)]

= (4a − b)(16a² + 4ab + b²)

∴  64a³ − b³ = (4a − b)(16a² + 4ab + b²)

Question: 6

Find the value

x³/216 − 8y³

Solution:

= (x/6 − 2y)((x/6)² + x/6 × 2y + (2y)²)

∴ [x³ − y³ = (x − y)(x² + xy + y²)]

= (x/6 − 2y)(x²/36 + xy/3 + 4y²)

∴ x³/216 − 8y³ = (x/6 − 2y)(x²/36 + xy/3 + 4y²)

Question: 7

Find the value

10x⁴y − 10xy⁴

Solution:

= 10xy(x³ − y³)

= 10xy(x − y)(x² + xy + y²)

∴ [x³ − y³ = (x − y)(x² + xy + y²)]

∴ 10x⁴y − 10xy⁴ = 10xy(x − y)(x² + xy + y²)

Question: 8

Find the value

54x⁶y + 2x³y⁴

Solution:

= 2x³y(27x³ + y³)

= 2x³y((3x)³ + y³)

= 2x³y(3x + y)((3x)² − 3x × y + y²)

∴ [a³ + b³ = (a + b)(a² − ab + b²)]

= 2x³y(3x + y)(9x² − 3xy + y²)

∴  54x⁶y + 2x³y⁴ = 2x³y(3x + y)(9x² − 3xy + y²)

Question: 9

Find the value

32a³ + 108b³

Solution:

= 4(8a³ + 27b³)

= 4((2a)³ + (3b)³)

= 4[(2a + 3b)((2a)² − 2a × 3b + (3b)²

∴ [a³ + b³ = (a + b)(a² − ab + b²)]

= 4(2a + 3b)(4a² − 6ab + 9b²)

∴ 32a³ + 108b³ = 4(2a + 3b)(4a² − 6ab + 9b²)

Question: 10

Find the value

(a − 2b)³ − 512b³

Solution:

= (a − 2b)³ − (8b)³

= (a − 2b − 8b)((a − 2b)² + (a − 2b)8b + (8b)²)

∴ [a³ − b³ = (a − b)(a² + ab + b²)]

= (a − 10b)(a² + 4b² − 4ab + 8ab − 16b² + 64b²)

= (a − 10b)(a² + 52b² + 4ab)

∴ (a − 2b)³ − 512b³ = (a − 10b)(a² + 52b² + 4ab)

Question: 11

Find the value

(a + b)³ − 8(a − b)³

Solution:

= (a + b)³ − [2(a − b)]³

= (a + b)³ − [2a − 2b]³

= (a + b − (2a − 2b))((a + b)² + (a + b)(2a − 2b) + (2a − 2b)²)

∴ [a³ − b³ = (a − b)(a² + ab + b²)]

= (a + b − 2a + 2b)(a² + b² + 2ab + (a + b)(2a − 2b) + (2a − 2b)²)

= (a + b − 2a + 2b)(a² + b² + 2ab + 2a² − 2ab + 2ab − 2b² + (2a − 2b)²)

= (3b − a)(3a² + 2ab − b² + (2a − 2b)²)

= (3b − a)(3a² + 2ab − b² + 4a² + 4b² − 8ab)

= (3b − a)(3a² + 4a² − b² + 4b² − 8ab + 2ab)

= (3b − a)(7a² +3b² − 6ab)

∴ (a + b)³ − 8(a − b)³ = (3b − a)(7a² + 3b² − 6ab)

Question: 12

Find the value

(x + 2)³ + (x − 2)³

Solution:

= (x + 2 + x − 2)((x + 2)² − (x + 2)(x − 2) + (x − 2)²)

∴ [a³ + b³ = (a + b)(a² − ab + b²)]

= 2x(x² + 4x + 4 − (x + 2)(x − 2) + x² − 4x + 4)

= 2x(2x² + 8 − (x² − 2²))

[∴ (a + b)(a − b) = a² − b²]

= 2x(2x² + 8 − x² + 4)

= 2x(x² + 12)

∴ (x + 2)³ + (x − 2)³ = 2x(x² + 12)

Question: 13

Find the value

8x²y³ − x⁵

Solution:

= x²((2y)³ − x³)

= x²(2y − x)((2y)² + 2y × x + x²)     

[∴ x³ − y³ = (x − y)(x² + xy + y²)]

= x²(2y − x)(4y² + 2xy + x²)

∴ 8x²y³ − x⁵ = x²(2y − x)(4y² + 2xy + x²)

Question: 14

Find the value

1029 - 3x³

Solution:

= 3(343 − x³)

= 3((7)³ − x³)

= 3(7 − x)(72 + 7x + x²) 

[∴ a³ − b³ = (a − b)(a² + ab + b²)]

= 3(7 − x)(49 + 7x + x²)

∴ 1029 - 3x³ = 3(7 − x)(49 + 7x + x²)

Question: 15

Find the value

x⁶ + y⁶

Solution:

= (x²)³ + (y²)³

= (x² + y²)((x²)² − x²y² + (y²)²)

= (x² + y²)(x⁴ − x²y² + y⁴) 

[∴ a³ + b³ = (a + b)(a² − ab + b²)]

∴ x⁶ + y⁶ = (x² + y²)(x⁴ − x²y² + y⁴)

Question: 16

Find the value

x³y³ + 1

Solution:

= (xy)³ + 1³

= (xy + 1)((xy)² + xy + 1²)

[∴ x³ + y³ = (x + y)(x² − xy + y²)]

= (xy + 1)(x²y² − xy + 1)

∴ x³y³ + 1 = (xy + 1)(x²y² − xy + 1)

Question: 17

Find the value

x⁴y⁴ − xy

Solution:

= xy(x³y³ − 1)

= xy((xy)³ − 1³)

= xy(xy − 1)((xy)² + xy × 1 + 12)

∴ [x³ − y³ = (x − y)(x² + xy + y²)]

= xy(xy − 1)(x²y² + xy + 1)

∴  x⁴y⁴ − xy = xy(xy − 1)(x²y² + xy + 1)

Question: 18

Find the value

a¹² + b¹²

Solution:

= (a⁴)³ + (b⁴)³

= (a⁴ + b⁴)((a⁴)² − a⁴ × b⁴ + (b⁴)²)

∴ [a³ + b³ = (a + b)(a² − ab + b²)]

= (a⁴ + b⁴)(a⁸ − a⁴b⁴ + b⁸)

∴ a¹² + b¹² = (a⁴ + b⁴)(a⁸ − a⁴b⁴ + b⁸)

Question: 19

Find the value

x³ + 6x² + 12x + 16

Solution:

= x³ + 6x² + 12x + 8 + 8

= x³ + 3 × x² × 2 + 3 × x × 2² + 2³ + 8

= (x + 2)³ + 8                                              

[∴ a³ + 3a²b + 3ab² + b³ = (a + b)³]

= (x + 2)³ + 23

= (x + 2 + 2)((x + 2)² − 2(x + 2) + 2²)

∴ [a³ + b³ = (a + b)(a² − ab + b²)]

= (x + 2 + 2)(x² + 4 + 4x − 2x − 4 + 4)

[∴ (a + b)² = a² + b² + 2ab]

= (x + 4)(x² + 4 + 2x)

∴ x³ + 6x² + 12x + 16 = (x + 4)(x² + 4 + 2x)

Question: 20

Find the value

a³ + b³ + a + b

Solution:

= (a³ + b³) + 1(a + b)

= (a + b)(a² − ab + b²) + 1(a + b)

[∴ a³ + b³ = (a + b)(a² − ab + b²)]

= (a + b)(a² − ab + b² + 1)

∴ a³ + b³ + a + b = (a + b)(a² − ab + b² + 1)

Question: 21

Find the value

a³ − 1/a³ − 2a + 2a

Solution:

= (a³ − 1/a³) − 2(a − 1/a)

= (a³ − (1/a)³) − 2(a − 1/a)

= (a − 1/a)(a² + a × 1/a + (1/a)²) − 2(a − 1/a)

[∴ a³ − b³ = (a − b)(a² + ab + b²)]

= (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)

∴ a³ − 1/a³ − 2a + 2a = (a − 1/a)(a² + 1/a² − 1)

Question: 22

Find the value

a³ + 3a²b + 3ab² + b³ − 8

Solution:

= (a + b)³ − 8 

[∴ a³ + 3a²b + 3ab² + b³ = (a + b)³]

= (a + b)³ − 23

= (a + b − 2)((a + b)² + (a + b) × 2 + 2²)

= (a + b - 2)(a² + 2ab + b² + 2a + 2b + 4)

∴ a³ + 3a²b + 3ab² + b³ − 8 = (a + b - 2)(a² + 2ab + b² + 2a + 2b + 4)

Question: 23

Find the value

8a³ − b³ − 4ax + 2bx

Solution:

= (2a)³ − b³ − 2x(2a − b)

= (2a − b)((2a)² + 2a × b + b²) − 2x(2a − b) 

[∴ a³ − b³ = (a − b)(a² + ab + b²)]

= (2a − b)(4a² + 2ab + b² − 2x)

∴ 8a³ − b³ − 4ax + 2bx = (2a − b)(4a² + 2ab + b² − 2x)

Question: 24

Find the value

Solution:

∴ [a³ + b³ = (a + b)(a² − ab + b²)]

= (173 + 127)

= 300

[∴ a³ − b³ = (a − b)(a² + ab + b²)]

= (1.2 − 0.2)

= 1.0

[∴ a³ − b³ = (a − b)(a² + ab + b²)]

= (155 - 55)

= 100