Factorizing a difference of two $n$-th powers

Question

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 $1$?

Answer 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.

Answer 2

Fix $b$.

Let $$P(X):=X^n-b^n-\displaystyle\prod_{j=1}^{n}(X-\omega^j b) \,.$$

Then $P(X)$ is a polynomial of degree $n-1$ and $P(\omega^j b)=0$ for all $1 \leq j \leq n$.

Thus $P(X)=0$ for all $X$, and in particular $P(a)=0$.