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. 
 A: ${x^3} - 1=(x-1)(x^2+x+1)$
${x^3} + 1=(x+1)(x^2-x+1)$
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 $n=0\cdots 5$,$w=\frac{\pi}{3}$ .
A: In general, we have:
$$x^n - 1 = (x-1)(x^{n-1}+x^{n-2}+\ldots+ 1).$$
It is a particular instance of: $$a^n-b^n = (a-b)(a^{n-1}+a^{n-2}b+\ldots+ab^{n-2}+b^{n-1}).$$

Think of the geometric progression $(a_1, \ldots, a_n,\ldots)$, with common ratio $q$. Then: $$a_1+\ldots +a_n = a_1 \frac{q^n-1}{q-1}.$$
This gives: $$1+x+\ldots+x^{n-1} = \frac{x^n-1}{x-1}.$$
A: The linear factors always include complex numbers as soon as $n>2$.  They always pair up into real quadratic factors $(x^2-2x\cos(2k\pi/n)+1)$
A: $$x^3 - 1 = (x-1)(x^2 + x + 1)$$
$$x^3 + 1 = (x+1)(x^2-x + 1)$$
A: By the Rational Root Theorem, the only possible rational roots of $x^3 - 1$ are $\pm 1$. Substituting gives that $x = 1$ is a root, so that $x - 1$ is a factor of $x^3 - 1$. Then polynomial division gives
$$x^3 - 1 = (x - 1) (x^2 + x + 1).$$
Similarly,
$$x^3 + 1 = (x + 1) (x^2 - x + 1).$$
The discriminant of $x^2 \pm x + 1$ is $-3 < 0$, so the quadratic factors do not factor over $\mathbb{R}$, so this furnishes the complete factorization of $x^6 - 1$ into polynomials irreducible over $\mathbb{R}$.
If you want a factorization over $\mathbb{C}$ instead, you can use that $$x^m - 1 = \prod_{i = 1}^m (x - \zeta_i),$$ where $\zeta_i$ varies over the $m$th roots of unity.
