Difference of Two Squares/Algebraic Proof 2

Proof
This is a special case of Difference of Two Powers:


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

The result follows by setting $n = 2$.