How do you factor #125x^3+8g^3#?

1 Answer
Feb 13, 2017

#125x^3+8g^3=(5x+2g)(25x^2-10xg+4g^2)#

Explanation:

As #125x^3+8g^3=(5x)^3+(2g)^3#, it can be factorized using identity #x^3+y^3=(x+y)(x^2-xy+y^2)#

and #125x^3+8g^3=(5x)^3+(2g)^3#

= #(5x+2g)((5x)^2-(5x)xx(2g)+(2g)^2)#

= #(5x+2g)(25x^2-10xg+4g^2)#

For a proof of the identity see below.

#x^3+y^3#

= #x^3+x^2y-x^2y-xy^2+xy^2+y^3#

= #x^2(x+y)-xy(x+y)+y^2(x+y)#

= #(x+y)(x^2-xy+y^2)#