How to factor $x-y$ out of $x^n - y^n$ for some arbitrary $n\in \mathbb{N}$

Question

As the title states. I don't quite understand how this factoring would work even though I'm fairly certain it can be done. Additionally, a follow up would be how does factoring $x^{n-1} - y^{n-1}$ out of $x^n - y^n$ work.

Answer 1

$$\begin{align} x^n - y^n &= (x-y)(x^{n-1} + x^{n-2}y + \cdots + xy^{n-2} + y^{n-1}) \\ &= (x-y)\sum_{k=0}^{n-1} x^{n-1-k}y^k \end{align}$$ generalizing the identity $x^3 - y^3 = (x-y)(x^2 + xy + y^2)$.

Answer 2

Also not much but a side-comment to this. For $n$ being even. We have $$x^n-y^n=\left(x-y\right)\left(x+y\right)\sum_{k=0}^{n/2-1}x^{\left(n-2-2k\right)}y^{2k}$$ For example, $$x^6-y^6=\left(x-y\right)\left(x+y\right)\left(x^{4}+x^{2}y^{2}+y^{4}\right)$$