Question

How do you factor `125x^6 - y^6`?

Answer 1

`125x^6-y^6=(sqrt(5)x-y)(5x^2+sqrt(5)xy+y^2)(sqrt(5)x+y)(5x^2-sqrt(5)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)`

Hence we find:

`125x^6-y^6`

`=(sqrt(125)x^3)^2-(y^3)^2`

`=(sqrt(125)x^3-y^3)(sqrt(125)x^3+y^3)`

`=((sqrt(5)x)^3-y^3)((sqrt(5)x)^3+y^3)`

`=(sqrt(5)x-y)((sqrt(5)x)^2+(sqrt(5)x)y+y^2)(sqrt(5)x+y)((sqrt(5)x)^2-(sqrt(5)x)y+y^2)`

`=(sqrt(5)x-y)(5x^2+sqrt(5)xy+y^2)(sqrt(5)x+y)(5x^2-sqrt(5)xy+y^2)`

There are no simpler factors with Real coefficients.

If you allow Complex coefficients then you can factor this further as:

`=(sqrt(5)x-y)(sqrt(5)x-omega y)(sqrt(5)x-omega^2 y)(sqrt(5)x+y)(sqrt(5)x+omega y)(sqrt(5)x+omega^2 y)`

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