Question

How do you factor completely `x^6 - y^6`?

Answer 1

Using the identity `a^2-b^2=(a+b)*(a-b)` we have that

`x^6-y^6=(x^3)^2-(y^3)^2=(x^3-y^3)*(x^3+y^3)`

Now we know that

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

and

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

So finally is

`x^6-y^6=(x-y)*(x+y)*(x^2-xy+y^2)*(x^2+xy+y^2)`

Answer 2

`x^6-y^6 = (x-y)(x^2+xy+y^2)(x+y)(x^2-xy+y^2)`

Explanation:

The difference of squares identity can be written:

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

The difference of cubes identity can be written:

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

The sum of cubes identity can be written:

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

So:

`x^6-y^6 = (x^3)^2-(y^3)^2 = (x^3-y^3)(x^3+y^3)`

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

If we allow Complex coefficients, then this reduces into linear factors:

`=(x-y)(x-omega y)(x-omega^2 y)(x+y)(x+omega y)(x+omega^2 y)`

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