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

##### 2 Answers

Using the identity

Now we know that

and

So finally is

#### 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