Question

How do you factor completely `64 - 27y^3`?

Answer 1

`64-27y^3 = (4-y)(16+4y+y^2)`

Explanation:

The difference of cubes identity can be written:

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

Note that `64=4^3` and `27y^3 = (3y)^3` are both perfect cubes, so we find:

`64-27y^3`

`=4^3-(3y)^3`

`= (4-y)(4^2+4y+y^2)`

`= (4-y)(16+4y+y^2)`

The remaining quadratic has negative discriminant. It cannot be factored further with Real coefficients. If we allow Complex coefficients then we can factor it completely:

`= (4-y)(4-omega y)(4-omega^2 y)`

where `omega = -1/2+sqrt(3)/2i` is the primitive Complex cube root of `1`.