# For MrCargoPants

Prove $$a^n - b^n = (a - b)\left(a^{n - 1} + a^{n - 2}b + \ldots + ab^{n - 2} + b^{n - 1}\right)$$

Here's the solution: First, I simply distribute $$a - b$$ over the complicated bit

\begin{align} & (a - b)\left(a^{n - 1} + a^{n - 2}b + \ldots + ab^{n - 2} + b^{n - 1}\right) \\ =\;& a\left(a^{n - 1} + a^{n - 2}b + \ldots + ab^{n - 2} + b^{n - 1}\right) \\ -\;& b\left(a^{n - 1} + a^{n - 2}b + \ldots + ab^{n - 2} + b^{n - 1}\right) \end{align}

I will now expand each term on the right-hand side

\begin{align} & a\left(a^{n - 1} + a^{n - 2}b + \ldots + ab^{n - 2} + b^{n - 1}\right) \\ =\;& a^n + a^{n - 1}b + \ldots + a^2b^{n - 2} + ab^{n - 1} \\[1em] & -b\left(a^{n - 1} + a^{n - 2}b + \ldots + ab^{n - 2} + b^{n - 1}\right) \\ =\;& -a^{n - 1}b + a^{n - 2}b^2 - \ldots - ab^{n - 1} - b^n \end{align}

Now I will add those together

\begin{align} & a^n + a^{n - 1}b + \ldots + a^2b^{n - 2} + ab^{n - 1} \\ -\;& a^{n - 1}b + a^{n - 2}b^2 - \ldots - ab^{n - 1} - b^n \end{align}

Notice that we have $$a^{n - 1}b$$ and $$-a^{n - 1}b,$$ and likewise for all terms except $$a^n$$ and $$b^n.$$ Thus

$$(a - b)\left(a^{n - 1} + a^{n - 2}b + \ldots + ab^{n - 2} + b^{n - 1}\right) = a^n - b^n$$