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

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


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

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

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

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

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

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

Taking the first term (where $$j = 0$$) out of the summation: $$b \circ S_n = \sum_{j=0}^{n-1} a^j \circ b^{n-j} = b^n + \sum_{j=1}^{n-1} a^{n-j} \circ b^j$$. This is equal to $$b^n + \sum_{j=1}^{n-1} a^j \circ b^{n-j}$$ by permutation of indices.

So:

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