Question

How do you factor `64x^6 -1`?

Answer 1

Use some standard identities to find:

`64x^6-1=(2x-1)(4x^2+2x+1)(2x+1)(4x^2-2x+1)`

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)`

We find:

`64x^6-1`

`=(8x^3)^2-1^2`

`=(8x^3-1)(8x^3+1)`

`=((2x)^3-1^3)((2x)^3+1^3)`

`=(2x-1)((2x)^2+2x+1)(2x+1)((2x)^2-2x+1)`

`=(2x-1)(4x^2+2x+1)(2x+1)(4x^2-2x+1)`

This is as far as we can go with Real coefficients, but if you allow Complex coefficients then this factors further as:

`=(2x-1)(2x-omega)(2x-omega^2)(2x+1)(2x+omega)(2x+omega^2)`

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