**Question 9. Verify:**

**(i) x ^{3} + y^{3} = (x + y) (x^{2} – xy + y^{2})**

**Solution:**

Formula (x + y)

^{3}= x^{3}+ y^{3}+ 3xy(x + y)x

^{3}+ y^{3}= (x + y)^{3}– 3xy(x + y)x

^{3}+ y^{3}= (x + y) [(x + y)^{2}– 3xy]

x

^{3}+ y^{3}= (x + y) [(x^{2}+ y^{2}+ 2xy) – 3xy]Therefore, x

^{3}+ y^{3}= (x + y) (x^{2}+ y^{2 }– xy)

**(ii) x ^{3} – y^{3} = (x – y) (x^{2} + xy + y^{2}) **

**Solution:**

Formula, (x – y)

^{3}= x^{3}– y^{3}– 3xy(x – y)x

^{3}− y^{3}= (x – y)^{3}+ 3xy(x – y)x

^{3}− y^{3}= (x – y) [(x – y)^{2}+ 3xy]x

^{3}− y^{3}= (x – y) [(x^{2 }+ y^{2 }– 2xy) + 3xy]

Therefore, x

^{3}+ y^{3}= (x – y) (x^{2}+ y^{2}+ xy)

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

**(i) 27y ^{3} + 125z^{3}**

**Solution:**

27y

^{3}+ 125z^{3}can also be written as (3y)^{3}+ (5z)^{3}27y

^{3}+ 125z^{3}= (3y)^{3}+ (5z)^{3}Formula x

^{3}+ y^{3}= (x + y) (x^{2}– xy + y^{2})27y

^{3}+ 125z^{3}= (3y)^{3}+ (5z)^{3}= (3y + 5z) [(3y)

^{2}– (3y)(5z) + (5z)^{2}]= (3y + 5z) (9y

^{2}– 15yz + 25z^{2})

**(ii) 64m ^{3} – 343n^{3}**

**Solution:**

64m

^{3}– 343n^{3}can also be written as (4m)^{3}– (7n)^{3}64m

^{3}– 343n^{3}= (4m)3 – (7n)^{3}Formula x

^{3}– y^{3}= (x – y) (x^{2}+ xy + y^{2})64m

^{3}– 343n^{3}= (4m)^{3}– (7n)^{3}= (4m – 7n) [(4m)

^{2}+ (4m)(7n) + (7n)^{2}]

**Question 11. Factorise: 27x**^{3} + y^{3} + z^{3} – 9xyz

^{3}+ y

^{3}+ z

^{3}– 9xyz

**Solution:**

27x

^{3}+ y^{3}+ z^{3}– 9xyz can also be written as (3x)^{3}+ y^{3}+ z^{3}– 3(3x)(y)(z)27x

^{3}+ y^{3}+ z^{3}– 9xyz = (3x)^{3}+ y^{3}+ z^{3}– 3(3x)(y)(z)Formula, x

^{3}+ y^{3}+ z^{3}– 3xyz = (x + y + z) (x^{2}+ y^{2}+ z^{2}– xy – yz – zx)27x

^{3}+ y^{3}+ z^{3}– 9xyz = (3x)^{3}+ y^{3}+ z^{3}– 3(3x)(y)(z)= (3x + y + z) [(3x)

^{2}+ y^{2}+ z^{2}– 3xy – yz – 3xz]= (3x + y + z) (9x

^{2}+ y^{2}+ z^{2}– 3xy – yz – 3xz)

**Question 12. Verify that: x**^{3} + y^{3} + z^{3} – 3xyz = (1/2) (x + y + z) [(x – y)^{2} + (y – z)^{2} + (z – x)^{2}]

^{3}+ y

^{3}+ z

^{3}– 3xyz = (1/2) (x + y + z) [(x – y)

^{2}+ (y – z)

^{2}+ (z – x)

^{2}]

**Solution:**

Formula, x

^{3}+ y^{3}+ z^{3 }− 3xyz = (x + y + z)(x^{2}+ y^{2}+ z^{2}– xy – yz – xz)Multiplying by 2 and dividing by 2

= (1/2) (x + y + z) [2(x

^{2}+ y^{2}+ z^{2}– xy – yz – xz)]= (1/2) (x + y + z) (2x

^{2}+ 2y^{2}+ 2z^{2}– 2xy – 2yz – 2xz)= (1/2) (x + y + z) [(x

^{2}+ y^{2}− 2xy) + (y^{2}+ z^{2}– 2yz) + (x^{2}+ z^{2}– 2xz)]= (1/2) (x + y + z) [(x – y)

^{2}+ (y – z)^{2}+ (z – x)^{2}]Therefore, x

^{3}+ y^{3}+ z^{3}– 3xyz = (1/2) (x + y + z) [(x – y)^{2}+ (y – z)^{2}+ (z – x)^{2}]

**Question 13. If x + y + z = 0, show that x**^{3} + y^{3} + z^{3} = 3xyz.

^{3}+ y

^{3}+ z

^{3}= 3xyz.

**Solution:**

Formula, x

^{3}+ y^{3}+ z^{3}– 3xyz = (x + y + z) (x^{2}+ y^{2}+ z^{2}– xy – yz – xz)Given, (x + y + z) = 0,

Then, x

^{3}+ y^{3}+ z^{3}– 3xyz = (0) (x^{2}+ y^{2}+ z^{2}– xy – yz – xz)x

^{3}+ y^{3}+ z^{3}– 3xyz = 0Therefore, x

^{3}+ y^{3}+ z^{3}= 3xyz

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

**(i) (−12) ^{3 }+ (7)^{3 }+ (5)^{3}**

**Solution:**

Let,

x = −12

y = 7

z = 5

We know that if x + y + z = 0, then x

^{3}+ y^{3}+ z^{3}= 3xyz.and we have −12 + 7 + 5 = 0

Therefore, (−12)

^{3}+ (7)^{3}+ (5)^{3}= 3xyz= 3 × -12 × 7 × 5

= -1260

**(ii) (28) ^{3} + (−15)^{3} + (−13)^{3}**

**Solution:**

Let,

x = 28

y = −15

z = −13

We know that if x + y + z = 0, then x

^{3}+ y^{3}+ z^{3}= 3xyz.and we have, x + y + z = 28 – 15 – 13 = 0

Therefore, (28)

^{3}+ (−15)^{3}+ (−13)^{3}= 3xyz= 3 (28) (−15) (−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 : 25a ^{2} – 35a + 12**

**Solution:**

Using splitting the middle term method,

25a

^{2}– 35a + 1225a

^{2}– 35a + 12 = 25a^{2}– 15a − 20a + 12= 5a(5a – 3) – 4(5a – 3)

= (5a – 4) (5a – 3)

Possible expression for length & breadth is = (5a – 4) & (5a – 3)

**(ii) Area : 35y ^{2} + 13y – 12**

**Solution:**

Using the splitting the middle term method,

35y

^{2}+ 13y – 12 = 35y^{2}– 15y + 28y – 12= 5y(7y – 3) + 4(7y – 3)

= (5y + 4) (7y – 3)

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

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

**(i) Volume : 3x ^{2} – 12x**

**Solution:**

3x

^{2 }– 12x can also be written as 3x(x – 4)= (3) (x) (x – 4)

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

**(ii) Volume: 12ky ^{2} + 8ky – 20k**

**Solution:**

12ky

^{2}+ 8ky – 20k can also be written as 4k (3y^{2}+ 2y – 5)12ky

^{2}+ 8ky– 20k = 4k(3y^{2}+ 2y – 5)Using splitting the middle term method.

= 4k (3y2 + 5y – 3y – 5)

= 4k [y(3y + 5) – 1(3y + 5)]

= 4k (3y + 5) (y – 1)

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