Question

How do you factor completely `x^3 + y^3`?

Answer 1

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

`color(white)(x^3+y^3) = (x+y)(x-(1/2+sqrt(3)/2i)y)(x-(1/2-sqrt(3)/2i)y)`

Explanation:

Given:

`x^3+y^3`

Note that if `y = -x` then this is zero. So we can deduce that `(x+y)` is a factor and separate it out:

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

We can calculate the discriminant for the remaining homogeneous quadratic in `x` and `y` just like we would for a quadratic in a single variable:

`x^2-xy+y^2`

is in standard form:

`ax^2+bxy+cy^2`

with `a=1`, `b=-1` and `c=1`

This has discriminant `Delta` given by the formula:

`Delta = b^2-4ac = color(blue)(1)^2-4(color(blue)(1))(color(blue)(1)) = -3`

Since `Delta < 0`, this quadratic has no linear factors with real coefficients.

We can factor it with complex coefficients by completing the square and using `i^2=-1` as follows:

`x^2-xy+y^2 = (x-1/2y)^2+3/4y^2`

`color(white)(x^2-xy+y^2) = (x-1/2y)^2+(sqrt(3)/2 y)^2`

`color(white)(x^2-xy+y^2) = (x-1/2y)^2-(sqrt(3)/2 y i)^2`

`color(white)(x^2-xy+y^2) = ((x-1/2y)-sqrt(3)/2i y)((x-1/2y)+sqrt(3)/2i y)`

`color(white)(x^2-xy+y^2) = (x-(1/2+sqrt(3)/2i)y)(x-(1/2-sqrt(3)/2i)y)`

So:

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

`color(white)(x^3+y^3) = (x+y)(x-(1/2+sqrt(3)/2i)y)(x-(1/2-sqrt(3)/2i)y)`