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# Class 9 NCERT Solutions – Chapter 2 Polynomials – Exercise 2.5 | Set 2

• Last Updated : 05 Apr, 2021

### Question 9. Verify:

(i) x3 + y3 = (x + y) (x2 – xy + y2)

Solution:

Formula (x + y)3 = x3 + y3 + 3xy(x + y)

x3 + y3 = (x + y)3 – 3xy(x + y)

x3 + y3 = (x + y) [(x + y)2 – 3xy]

x3 + y3 = (x + y) [(x2 + y2 + 2xy) – 3xy]

Therefore, x3 + y3 = (x + y) (x2 + y2 – xy)

(ii) x3 – y3 = (x – y) (x2 + xy + y2)

Solution:

Formula, (x – y)3 = x3 – y3 – 3xy(x – y)

x3 âˆ’ y3 = (x – y)3 + 3xy(x – y)

x3âˆ’ y3 = (x – y) [(x – y)2 + 3xy]

x3 âˆ’ y3 = (x – y) [(x2 + y2 â€“ 2xy) + 3xy]

Therefore, x3 + y3 = (x â€“ y) (x2 + y2 + xy)

### Question 10. Factorize each of the following:

(i) 27y3 + 125z3

Solution:

27y3 + 125z3 can also be written as (3y)3 + (5z)3

27y3 + 125z3 = (3y)3 + (5z)3

Formula x3 + y3 = (x + y) (x2 â€“ xy + y2)

27y3 + 125z3 = (3y)3 + (5z)3

= (3y + 5z) [(3y)2 â€“ (3y)(5z) + (5z)2]

= (3y + 5z) (9y2 â€“ 15yz + 25z2)

(ii) 64m3 – 343n3

Solution:

64m3 – 343n3 can also be written as (4m)3 – (7n)3

64m3 – 343n3 = (4m)3 – (7n)3

Formula x3 – y3 = (x – y) (x2 + xy + y2)

64m3 – 343n3 = (4m)3 – (7n)3

= (4m – 7n) [(4m)2 + (4m)(7n) + (7n)2]

### Question 11. Factorise: 27x3 + y3 + z3 â€“ 9xyz

Solution:

27x3 + y3 + z3 – 9xyz can also be written as (3x)3 + y3 + z3 – 3(3x)(y)(z)

27x3 + y3 + z3 â€“ 9xyz  = (3x)3 + y3 + z3 â€“ 3(3x)(y)(z)

Formula, x3 + y3 + z3 â€“ 3xyz = (x + y + z) (x2 + y2 + z2 â€“ xy  â€“ yz â€“ zx)

27x3 + y3 + z3 â€“ 9xyz  = (3x)3 + y3 + z3 â€“ 3(3x)(y)(z)

= (3x + y + z) [(3x)2 + y2 + z2 â€“ 3xy â€“ yz â€“ 3xz]

= (3x + y + z) (9x2 + y2 + z2 â€“ 3xy â€“ yz â€“ 3xz)

### Question 12. Verify that: x3 + y3 + z3 â€“ 3xyz  = (1/2) (x + y + z) [(x â€“ y)2 + (y â€“ z)2 + (z â€“ x)2]

Solution:

Formula, x3 + y3 + z3 âˆ’ 3xyz = (x + y + z)(x2 + y2 + z2 â€“ xy â€“ yz â€“ xz)

Multiplying by 2 and dividing by 2

= (1/2) (x + y + z) [2(x2 + y2 + z2 â€“ xy â€“ yz â€“ xz)]

= (1/2) (x + y + z) (2x2 + 2y2 + 2z2 â€“ 2xy â€“ 2yz â€“ 2xz)

= (1/2) (x + y + z) [(x2 + y2 âˆ’ 2xy) + (y2 + z2 â€“ 2yz) + (x2 + z2 â€“ 2xz)]

= (1/2) (x + y + z) [(x â€“ y)2 + (y â€“ z)2 + (z â€“ x)2]

Therefore, x3 + y3 + z3 â€“ 3xyz  = (1/2) (x + y + z) [(x â€“ y)2 + (y â€“ z)2 + (z â€“ x)2]

### Question 13. If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.

Solution:

Formula, x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 â€“ xy â€“ yz â€“ xz)

Given, (x + y + z) = 0,

Then, x3 + y3 + z3 – 3xyz = (0) (x2 + y2 + z2 â€“ xy â€“ yz â€“ xz)

x3 + y3 + z3 â€“ 3xyz = 0

Therefore, x3 + y3 + z3 = 3xyz

### Question 14. Without actually calculating the cubes, find the value of each of the following:

(i) (âˆ’12)3 + (7)3 + (5)3

Solution:

Let,

x = âˆ’12

y = 7

z = 5

We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz.

and we have âˆ’12 + 7 + 5 = 0

Therefore, (âˆ’12)3 + (7)3 + (5)3 = 3xyz

= 3 Ã— -12 Ã— 7 Ã— 5

= -1260

(ii) (28)3 + (âˆ’15)3 + (âˆ’13)3

Solution:

Let,

x = 28

y = âˆ’15

z = âˆ’13

We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz.

and we have, x + y + z = 28 â€“ 15 â€“ 13 = 0

Therefore, (28)3 + (âˆ’15)3 + (âˆ’13)3 = 3xyz

= 3 (28) (âˆ’15) (âˆ’13)

= 16380

### Question 15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:

(i) Area : 25a2 â€“ 35a + 12

Solution:

Using splitting the middle term method,

25a2 â€“ 35a + 12

25a2 â€“ 35a + 12 = 25a2 â€“ 15a âˆ’ 20a + 12

= 5a(5a â€“ 3) â€“ 4(5a â€“ 3)

= (5a â€“ 4) (5a â€“ 3)

Possible expression for length & breadth is  = (5a â€“ 4)  & (5a  â€“ 3)

(ii) Area : 35y2 + 13y â€“ 12

Solution:

Using the splitting the middle term method,

35y2 + 13y â€“ 12 = 35y2 â€“ 15y + 28y â€“ 12

= 5y(7y â€“ 3) + 4(7y â€“ 3)

= (5y + 4) (7y â€“ 3)

Possible expression for length  & breadth is = (5y + 4) & (7y â€“ 3)

### Question 16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?

(i) Volume : 3x2 â€“ 12x

Solution:

3x2 â€“ 12x can also be written as 3x(x â€“ 4)

= (3) (x) (x â€“ 4)

Possible expression for length, breadth & height = 3, x & (x – 4)

(ii) Volume: 12ky2 + 8ky â€“ 20k

Solution:

12ky2 + 8ky â€“ 20k can also be written as 4k (3y2 + 2y â€“ 5)

12ky2 + 8kyâ€“ 20k = 4k(3y2 + 2y â€“ 5)

Using splitting the middle term method.

= 4k (3y2 + 5y â€“ 3y â€“ 5)

= 4k [y(3y + 5) â€“ 1(3y + 5)]

= 4k (3y + 5) (y â€“ 1)

Possible expression for length, breadth & height= 4k, (3y + 5) & (y – 1)

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