Class 9 Maths

# Polynomials

## Exercise 2.5 Part 10

Question: 9 – Verify:

(i) x^3 + y^3 = (x + y)(x^2 - xy + y^2)

Answer: RHS = (x + y)(x^2 - xy + y^2

= x^3 - x^2\y + xy^2 + yx^2 - xy^2 + y^3

= x^3 + y^3 = LHS proved

(ii) x^3 + y^3 = (x - y)(x^2 + xy + y^2)

Answer: RHS = (x - y)(x^2 + xy + y^2

= x^3 + x^2\y + xy^2 - yx^2 - xy^2 - y^3

= x^3 - y^3 = LHS Proved

Question: 10 – Factorise each of the following:

(i) 27y^3 + 125z^3

Answer: Given; 27y^3 + 125z^3

= (3y)^3 + (5z)^3

Using the identity x^3 + y^3 = (x + y)(x^2 - xy + y^2)

We get: 27y^3 + 125z^3

= (3y + 5z)[(3y)^2 - 3y\xx5z + (5z)^2]

= (3y + 5z)(9y^2 - 15yz + 25z^2)

(ii) 64m^3 - 343n^3

Answer: Given; 64m^3 - 343n^3

= (4m – 7n)[(4m)^2 + 4m\xx7n + (7n)^2]

= (4m – 7n)(16m^2 + 28mn + 49n^2)

Question: 11 – Factorise:

27x^3 + y^3 + z^3 - 9xy\z

Answer: Given; 27^3 + y^3 + z^3 - 9xy\z

= (3x)^3 + y^3 + z^3 - 3xx3xy\z

Using the identity x^3 + y^3 + z^3 - 3xy\z

= (x + y + z)(x^2 + y^2 + z^2 - xy – yz – xz)

We get: (3x + y + z)[(3x)^2 + y^2 + z^2 - 3xy – yz – 3xz]

= (3x + y + z)(9x^2 + y^2 + z^3 - 3xy – yz – 3xz)