Question

How do you factor `8a^3+27(x+y)^3`?

Answer 1

The answer is:

`(2a+3x+3y)(4a^2-6ax-6ay+9x^2+18xy+9y^2)`

If you write:

`8a^3+27(x+y)^3=(2a)^3+[3(x+y)]^3=(1)`

This is a sum of two cubic, and you can use this rule:

`a^3+b^3=(a+b)(a^2-ab+b^2)`.

So:

`(1)=[2a+3(x+y)][(2a)^2-(2a)*3(x+y)+3^2(x+y)^2]=`

`=(2a+3x+3y)(4a^2-6ax-6ay+9x^2+18xy+9y^2)`.

Answer 2

Temporarily simplify be replacing
`8a^3 = (2a)^3` with `p^3`
and
`27(x+y)^3 = (3(x+y))^3` with `q^3`

So `8a^3+27(x+y)^3`
becomes `p^3+q^3`
which has factors
`(p+q)(p^2-pq+q^2)`

Restoring the original values, this becomes
`(2a + 3(x+y))*(4a^2 - 6a(x+y) +9(x+y)^2)`

It might be tempting to attempt to further factor the right-most term but a quick check using the formula for roots
`(-b+-sqr(b^2-4ac))/2a` reveals that no Real roots are possible
[be careful not to confuse the `a` in this general formula with the `a` in the example expression]