Factorizing a difference of two $n$-th powers How can be proved that
$$a^n-b^n=\displaystyle\prod_{j=1}^{n}(a-\omega^j b)$$
where $\omega=e^{\frac{2\pi i}{n}}$ is a primitive $n$-th root of $1$?

Hint: $$a^n-b^n = b^n((a/b)^n-1).$$
Now you can represent a polynomial as a product of terms $(x-x_i)$ where $x_i$ are its roots.
