Factorization of $x^6 - 1$ I started by intuition since I'm familiar with the formula $a^2-b^2 = (a-b)(a+b)$. So in our case $$x^6 - 1 = ({x^3} - 1)({x^3}+1)$$
How should I proceed? I assume there's some sort of algorithm to keep the process till you reach a form of irreducible linear polynomials. 

The linear factors always include complex numbers as soon as $n>2$.  They always pair up into real quadratic factors $(x^2-2x\cos(2k\pi/n)+1)$
