How to factorise $x^6 + 1$?

Question

How do I factorise this? I already said that

$$x^6 + 1 = 0 \implies (x^3)^2 + 1 = 0$$

so we then get

$$(x^3)^2 = - 1$$

and then there are no real roots. Similar thing happens if I try it as $(x^2)^3$.

How would I then go about factorising this? I know its possible as

$$(x^6 + 1) = (x^2 + 1)(x^4 - x^2 + 1).$$

Answer 1

You're on the right track with the $(x^3)^2$, ... thing. Use $(x^2)^3$ -- interpret it as a polynomial in $x^2$ (which will be cubic), as opposed to one in $x$. That is, write it as $u^3 + 1$ with $u = x^2$. Now factor that.

Answer 2

Hint $\ $ By the Factor Theorem $\rm\ z-c\mid z^3 - c^3.\ $ Let $\rm\ z = x^2,\,\ c = -1$.