Difference of Two Powers

Theorem
Let $\left({R, +, \circ}\right)$ be a commutative ring whose zero is $0_R$.

Let $a, b \in R$.

Let $n \in \N$ such that $n \ge 2$.

Then:
 * $\displaystyle a^n - b^n = \left({a - b}\right) \circ \left({a^{n-1} + a^{n-2} \circ b + a^{n-3} \circ b^2 + \ldots + a \circ b^{n-2} + b^{n-1}}\right) = \left({a - b}\right) \circ \sum_{j=0}^{n-1} a^{n-j-1} \circ b^j$

When $R$ is one of the standard sets of numbers, i.e. $\Z, \Q, \R$ etc., then this translates into:


 * $\displaystyle a^n - b^n = \left({a - b}\right) \left({a^{n-1} + a^{n-2} b + a^{n-3} b^2 + \ldots + a b^{n-2} + b^{n-1}}\right) = \left({a - b}\right) \sum_{j=0}^{n-1} a^{n-j-1} b^j$

Proof
Let $\displaystyle S_n = \sum_{j=0}^{n-1} a^{n-j-1} \circ b^j$.

This can also be written $\displaystyle S_n = \sum_{j=0}^{n-1} a^j \circ b^{n-j-1}$.

Consider $\displaystyle a \circ S_n = \sum_{j=0}^{n-1} a^{n-j} \circ b^j$.

Taking the first term (where $j = 0$) out of the summation, we get:
 * $\displaystyle a \circ S_n = \sum_{j=0}^{n-1} a^{n-j} \circ b^j = a^n + \sum_{j=1}^{n-1} a^{n-j} \circ b^j$

Similarly, consider $\displaystyle b \circ S_n = \sum_{j=0}^{n-1} a^j \circ b^{n-j}$.

Taking the first term (where $j = 0$) out of the summation:
 * $\displaystyle b \circ S_n = \sum_{j=0}^{n-1} a^j \circ b^{n-j} = b^n + \sum_{j=1}^{n-1} a^{n-j} \circ b^j$

This is equal to $\displaystyle b^n + \sum_{j=1}^{n-1} a^j \circ b^{n-j}$ by permutation of indices.

So: