Odd powers make $x+a$ a factor; even powers do not unless $2a^n$ vanishes.Suggest a slogan
Let \(R\) be a commutative ring, let \(a\in R\), and let \(n\ge1\). If \(n\) is odd, then in \(R[x]\) \[ x^n+a^n=(x+a)\sum_{k=0}^{n-1}(-a)^kx^{n-1-k}. \] If \(n\) is even and \(2a^n\ne0\) in \(R\), for example if \(R\) is a field of characteristic other than \(2\) and \(a\ne0\), then \(x+a\) does not divide \(x^n+a^n\) in \(R[x]\).
For odd \(n\), \((-a)^n=-a^n\), so the factorization is FF-ED3CD applied to \(-a\). Since \(x+a\) is monic, division by \(x+a\) in \(R[x]\) leaves the constant remainder \(f(-a)\) for every \(f\in R[x]\). For even \(n\) the remainder of \(x^n+a^n\) is \((-a)^n+a^n=2a^n\ne0\).
When does $x + a$ divide $x^n + a^n$, and what is the factorization?