Question

How do you factor `1-x^8` ?

Answer 1

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

Explanation:

Remember `(a^2-b^2)=(a-b)(a+b)`
`(1-x^8)=(1-x^4)(1+x^4)`
=`(1-x^2)(1+x^2)(1+x^4)`
=`(1-x)(1+x)(1+x^2)(1+x^4)`

Answer 2

`1-x^8 = (1-x)(1+x)(1+x^2)(1-sqrt(2)x+x^2)(1+sqrt(2)x+x^2)`

Explanation:

The difference of squares identity can be written:

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

Note also that:

`(a^2-kab+b^2)(a^2+kab+b^2) = a^4+(2-k^2)a^2b^2+b^4`

In particular, putting `k = sqrt(2)` we find:

`(a^2-sqrt(2)ab+b^2)(a^2+sqrt(2)ab+b^2) = a^4+b^4`

So we find:

`1 - x^8 = 1^2-(x^4)^2`

`color(white)(1-x^8) = (1-x^4)(1+x^4)`

`color(white)(1-x^8) = (1^2-(x^2)^2)(1^4+x^4)`

`color(white)(1-x^8) = (1-x^2)(1+x^2)(1^2-sqrt(2)x+x^2)(1^2+sqrt(2)x+x^2)`

`color(white)(1-x^8) = (1^2-x^2)(1+x^2)(1-sqrt(2)x+x^2)(1+sqrt(2)x+x^2)`

`color(white)(1-x^8) = (1-x)(1+x)(1+x^2)(1-sqrt(2)x+x^2)(1+sqrt(2)x+x^2)`

The remaining quadratic factors have no linear factors with Real coefficients.