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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


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