Question

How do you factor completely: `x^8-9`?

Answer 1

`x^8-9=(x-3^(1/4))(x+3^(1/4))(x-i3^(1/4))(x+i3^(1/4))(x-(1/sqrt(2)+i/sqrt(2))3^(1/4))(x+(1/sqrt(2)+i/sqrt(2))3^(1/4))(x-(1/sqrt(2)-i/sqrt(2))3^(1/4))(x+(1/sqrt(2)-i/sqrt(2))3^(1/4))`

Explanation:

Using the difference of squares factorisation (`a^2-b^2=(a-b)(a+b)`) you have:
`x^8-9=(x^4-3)(x^4+3)`

This is probably all they want but you can factor further allowing complex numbers:
`(x^4-3)(x^4+3)=`
`(x^2-3^(1/2))(x^2+3^(1/2))(x^2-i3^(1/2))(x^2+i3^(1/2))=`
`(x-3^(1/4))(x+3^(1/4))(x-i3^(1/4))(x+i3^(1/4))(x-(1/sqrt(2)+i/sqrt(2))3^(1/4))(x+(1/sqrt(2)+i/sqrt(2))3^(1/4))(x-(1/sqrt(2)-i/sqrt(2))3^(1/4))(x+(1/sqrt(2)-i/sqrt(2))3^(1/4))`

The 8 roots are the 8 solutions to: `x^8=9`

Answer 2

Factor `x^8 - 9`

Explanation:

`x^8 - 9 = (x^4 - 3)(x^4 + 3) =`

= `(x^2 - sqrt3)(x^2 + sqrt3)(x^4 + 3)`

= `(x - root(4)(3))(x + root(4)(3))(x^2 + sqrt3)(x^4 + 3)`