Question

How do you factor `72x^ { 3} - 125y ^ { 3}`?

Answer 1

the answer is `(root(3)72x-5y)(root(3)72x)^2+root(3)72x*5y+25y^2)`

Explanation:

You can use the identity
`a^3-b^3=(a-b)(a^2+ab+b^2)`
Applying this to our case
`a=root(3)72x` and `b=5y`
so `72x^3-125y^3=(root(3)72x-5y)((root(3)72x)^2+root(3)72x*5y+25y^2)`

Answer 2

`27x^3 -125y^3 = (3x-5y)(9x^2 +15xy +25y^2)`

Explanation:

The nature of the numbers and the powers makes me think that there is a typo in the question.

As there is already an answer, I will treat is as though it is an error and it was supposed to be

`27x^3 -125y^3`

This is an expression known as the difference of two cubes and it factors by a set rule:

recall: `(x^3 -y^3) = (x-y)(x^2 +xy + y^2)`

`27x^3 -125y^3 = (3x-5y)(9x^2 +15xy +25y^2)`