Factorization of $x^n - a^n$

A difference of $n$th powers always carries the linear factor $x-a$.Suggest a slogan

Let $R$ be a commutative ring, let $a\in R$, and let $n\ge1$. In $R[x]$, \[ x^n-a^n=(x-a)\sum_{k=0}^{n-1}a^kx^{n-1-k}. \]

Expanding, \[ (x-a)\sum_{k=0}^{n-1}a^kx^{n-1-k}=\sum_{k=0}^{n-1}a^kx^{n-k}-\sum_{k=1}^{n}a^kx^{n-k}, \] and the two sums cancel except for the terms $x^n$ and $-a^n$.