TITLE: Derivation of factorization of $a^n-b^n$
QUESTION [22 upvotes]: How does one prove that:
$$a^n-b^n=(a-b)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\dots+a^2b^{n-3}+ab^{n-2}+b^{n-1}\right)$$
Better yet, why is $a^n-b^n$ divisible by $a-b$? I would very much appreciate some help on this. Thanks

REPLY [2 votes]: $Q=\sum_{i=1}^{n}{a^{n-i}b^{i-1}}=a^{n-1}+a^{n-2}b+\dots+ab^{n-2}+b^{n-1}$
$a^n-b^n=(a-b)Q=(a)Q+Q(-b)$
$\begin{matrix}
(a)Q&=&a^n&+a^{n-1}b&+a^{n-2}b^2&+\dots&+a^2b^{n-2}&+ab^{n-1}&\\
+Q(-b)&=&&-a^{n-1}b&-a^{n-2}b^2&-\dots&-a^2b^{n-2}&-ab^{n-1}&-b^n\\
\end{matrix}$