ML Aggarwal Solutions Class 9 Math Fourth Chapter Factorisation Exercise 4.5

ML Aggarwal Solutions Class 9 Math 4th Chapter Factorisation Exercise 4.5

ML Aggarwal Understanding ICSE Mathematics Class 9 Solutions Fourth Chapter Factorisation Exercise 4.5. APC Solution Class 9 Exercise 4.5.

 

(1) (i) 8x3+ y3

Solution:

8x3 + y3

= (2x)3+ y3

= (2x + y) (2x)2 – 2x.y + y2)

= (2x + y) (4x2 – 2xy + y2)

 

(ii) 64x3 – 125y3

Solution:

64x3 – 125y3

= (4x)3 – (5y)3

= (4x – 5y) ((4x)2 + 4x.5y + (5y)2)

= (4x – 5y) (16x2 + 20xy + 25y2)

 

(2) (i) 64x3 + 1

Solution:

64x3 + 1

= (4x)3+ 13

= (4x + 1) (4x)2 – 4x.1 + 12)

= (4x + 1) (16x2 – 4x + 1)

 

(ii) 7a3 + 56b3

Solution:

7a3 + 56b3

= 7 (a3 + 8b3)

= 7 (a3 + (2b)3)

= 7 ((a + 2b) (a2 – a.2b + (2b)2)

= 7 (a + 2b) (a2 – 2ab + 4b2)

 

(3) (i) x6/343 + 343/x6

Solution:

x6/343 + 343/x6

= (x2/7)3 + (7/x2)3

= (x2/7 + 7/x2) ((x2/7)2 – x3/7. 7/x3 + (7/x2)2)

= (x2/7 + 7/x2) (x4/4 + 44/x)

 

(ii) 8x3 – 1/27y3

Solution:

8x3 – 1/27y3

= (2x)3 – (1/3y)3

= (2x – 1/3y) (2x)2 + 2x.1/3y + (1/3y)2)

= (2x – 1/3y) (4x2 + 2x/3y + 1/9y2)

 

(4) (i) x2 + x5

Solution:

x2 + x5

= x2 (1 + x3)

= x2 (1 + x) (12 – 1.x + x2)

= x2 (1 + x) (1 – x + x2)

 

(ii) 32x4 – 500x

Solution:

32x4 – 500x

= 4x (8x3 – 125)

= 4x ((2x)3 – 53)

= 4x ((2x – 5) ((2x)2 + 2x.5 + 52))

= 4x (2x – 5) (4x2 + 10x + 25)

 

(5) (i) 27x3y3 – 8

Solution:

27x3y3 – 8

= (3xy)3 – 23

= (3xy – 2) ((3xy)2 + 3xy.2 + 22)

= (3xy – 2) (9x2y2 + 6xy + 4)

 

(ii) 27 (x + y)3 + 8 (2x – y)3

Solution:

27 (x + y)3 + 8 (2x – y)3

Let, x + y = a, 2x – y = b

∴ 27 (a)3 + 8b3

= (3a)3 + (2b)3

= (3a + 2b) (3a)2 – 3a.2b + (2b)2)

= (3a + 2b) (9a2 – 6ab + 4b2)

Putting the value of a & b we get,

= (3 (x + y) + 2 (2x + y)) (9 (x + y)2 – 6 (x + y) (2x – y) + 4 (2x – y)2}

= (3x + 3y + 4x – 2y) (9 (x2 + 2xy + y2) – 6 (2x2 – xy + 2xy – y2) + 4 (4x2 – 4xy + y2)

= (7x + y) (9x2 + 18xy + 9y2 – 12x2 – 6xy + 6y2 + 16x2 – 16xy + 4y2)

= (7x + y) (13x2 – 4xy 19y2)

 

(6) (i) a3 + b3 + a + b

Solution:

a3 + b3 + a + b

= (a + b) (a2 – ab + b2) + 1 (a + b)

= (a + b) (a2 – ab + b2 + 1)

 

(ii) a3 – b3 – a + b

Solution:

a3 – b3 – a + b

= (a – b) (a2 + ab + b2) – 1 (a – b)

= (a + b) (a2 + ab + b2 – 1)

 

(7) (i) x3 + x + 2

Solution:

x3 + x + 2

= x3 + 1 + x + 1

= (x + 1) (x2 – x + 1) + 1 (x + 1)

= (x + 1) (x2 – x + 1 + 1)

= (x + 1) (x2 – x + 2)

 

(8) (i) x3 + 6x2 + 12x + 16

Solution:

x3 + 6x2 + 12x + 16

= x3 + 6x2 + 12x + 8 + 8

= x3 + 3x2 × 2 + 3x.22 + 23 + 8

= (x + 2)3 + 8

= (x + 2)3 + 23

= (x + 2 + 2) ((x + 2)2 – (x + 2) 2 + 22)

= (x + 4) (x2 + 4 – 2x – 4 + 4)

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

 

(ii) a3 – 3a2b + 3ab2 – 2b3

Solution:

a3 – 3a2b + 3ab2 – 2b3

= a3 – 3a2b + 3ab2 – b3 – b3

= (a – b)3 – b3

= (a – b – b)  (a3 + ab + b2)

= (a – 2b) (a2 + ab + b2)

 

(9) (i) 2a3 + 16b3 – 5a – 10b

Solution:

2a3 + 16b3 – 5a – 10b

= 2 (a3 + 8b3) – 5 (a + 2b)

= 2 (a3 + (2b)3) – 5 (a + 2b)

= 2 (a + 2b) (a2 – 2ab + 4b2) – 5 (a + b)

= 2 (a + 2b) (a2 – 2ab + 4b2 – 5)

 

(ii) a3 – 1/a3 – 2a + 2/a

Solution:

a3 – 1/a3 – 2a + 2/a

= a3 – 1/a3 – 2 (a – 1/a)

= (a – 1/a) (a2 + a .1/a + 1/a2) – 2 (a – 1/a)

= (a – 1/a) (a2 + 1 + 1/a2 – 2)

= (a – 1/a) (a2 + 1/a2 – 1)

 

(10) (i) a6 – b6

Solution:

a6 – b6

= (a3)2 – (b3)2

= (a3 + b3) (a3 – b3)

= (a + b) (a2 – ab + b2) (a – b) (a2 + ab + b2)

 

(ii) x6 – 1

Solution:

x6 – 1

= (x3)2 – 1

= (x3 + 1) (x3 – 1)

= (x + 1) (x2 – x + 1) (x – 1) (x2 + x + 1)

 

(11) (i) 64x6 – 729y6

Solution:

64x6 – 729y6

= (8x3)2 – (27y3)2

= (8x3 – 27y3) (8x3 + 27y3)

= ((2x)3 – (3y)3) (2x)3 + (3y)3)

= (2x – 3y) (4x2 + 6xy + 9y2) (2x + 3y) (4x2 – 6xy + 9y2)

 

(12) (i) 250 (a – b)3 + 2

Solution:

250 (a – b)3 + 2

= 2 (125 (a – b)3 + 1)

= 2 ((5 (a – b)) ((5 (a – b))2 + 5 (a – b).1 + 1)))

= 2 (5a – 5b) (25a2 – 5ab + 5a – 5b + 1)

= 2 (5a – 5b + 1) (25a2 – 5ab + 25b2 – 5ab + 1)

 

(ii) 32a2 x3 – 8b2x3 – 4a2 y3+ b2 y3

Solution:

32a2 x3 – 8b2x3 – 4a2 y3 + b2 y3

= 32a2 x3 – 4a2 y3 – 8b2 x3 + b2 y3

= 4a2 (8x3 – y3) – b2 (8x3 – y3)

= (4a2 – b2) ((2x)3 – y3)

= (2a + b) (2a – b) (2x – y) (4x2 + 2xy + y2)

 

(13) (i) x9 + y9

Solution:

x9 + y9

= (x3)3 + (y3)3

= (x3 + y3) (x6 – x3y3 + y6)

= (x + y) (x2 – 2y + y2) (x6 – x3y3 + y6)

 

(ii) x6 – 7x3 – 8

Solution:

x6 – 7x3 – 8

= x4 – 8x3 + x3 – 8

= x3 (x3 – 8) + 1 (x3 – 8)

= (x3 + 1) (x3 – 8)

= (x3 + 13) (x3 – 23)

= (x + 1) (x2 – x + 1) (x – 2) (x2 + 2x + 2)

Updated: June 17, 2022 — 5:05 pm

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