Question:
Let
\[x^{12} - 1 = p_1(x) p_2(x) \dotsm p_k(x),\]where each non-constant polynomial $p_i(x)$ is monic with integer coefficients, and cannot be factored further over the integers.  Find $k.$

Answer:
First, we can apply difference of squares, to get
\[x^{12} - 1 = (x^6 - 1)(x^6 + 1).\]We can apply difference of squares to $x^6 - 1$:
\[x^6 - 1 = (x^3 - 1)(x^3 + 1).\]These factor by difference of cubes and sum of cubes:
\[(x^3 - 1)(x^3 + 1) = (x - 1)(x^2 + x + 1)(x + 1)(x^2 - x + 1).\]Then by sum of cubes,
\[x^6 + 1 = (x^2 + 1)(x^4 - x^2 + 1).\]Thus, the full factorization over the integers is
\[x^{12} - 1 = (x - 1)(x^2 + x + 1)(x + 1)(x^2 - x + 1)(x^2 + 1)(x^4 - x^2 + 1),\]and there are $\boxed{6}$ factors.