Question

2. Factorize:
(a) 1-729m3
(b) 125x3 - 27y3
​

Answer 1

GIVEN :

Factorize:

(a)

(b) $125x^3 - 27y^3$

TO FIND :

The factors of the expressions

(a) $1-729m^3$

(b) $125x^3 - 27y^3$

SOLUTION :

Given that the expressions are :

(a) $1-729m^3$

(b) $125x^3 - 27y^3$

Solving (a) $1-729m^3$

$=1^3-9^3m^3$

By using the exponent property :

$a^mb^m=(ab)^m$

$=1^3-(9m)^3$

By using the algebraic identity :

$a^3-b^3=(a-b)(a^2+ab+b^2)$

Here a=1 and b=9m

$=(1-9m)(1^2+1(9m)+(9m)^2)$

$=(1-9m)(1+9m+9^2m^2)$

$=(1-9m)(1+9m+81m^2)$

The given expression factorised into the factors$1-729m^3=(1-9m)(1+9m+81m^2)$

Solving (b) $125x^3-27y^3$

$=5^3x^3-3^3y^3$

By using the exponent property :

$a^mb^m=(ab)^m$

$=(5x)^3-(3y)^3$

By using the algebraic identity :

$a^3-b^3=(a-b)(a^2+ab+b^2)$

Here a = 5x and b=3y

$=(5x-3y)((5x)^2+5x(3y)+(3y)^2)$

$=(5x-3y)(5^2x^2+15xy+3^2y^2)$

$=(5x-3y)(25x^2+15xy+9y^2)$

The given expression factorised into the factors$125x^3-27y^3=(5x-3y)(25x^2+15xy+9y^2)$