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. 

${x^3} - 1=(x-1)(x^2+x+1)$
${x^3} + 1=(x+1)(x^2-x+1)$
The last quadratic term is already irreducible in $\mathbb{R}$.
If you mean irreducible in $\mathbb{C}$, you may split it into product of factor $(x-e^{inw})$, where $n=0\cdots 5$,$w=\frac{\pi}{3}$ .
