Difference of Squares of Sum and Difference

Theorem

$\forall a, b \in \R: \left({a + b}\right)^2 - \left({a - b}\right)^2 = 4 a b$

Algebraic Proof

 $\displaystyle$  $\displaystyle \left({a + b}\right)^2 - \left({a - b}\right)^2$ $\displaystyle$ $=$ $\displaystyle \left({a^2 + 2 a b + b^2}\right) - \left({a^2 - 2 a b + b^2}\right)$ Square of Sum and Square of Difference $\displaystyle$ $=$ $\displaystyle a^2 + 2 a b + b^2 - a^2 + 2 a b - b^2$ $\displaystyle$ $=$ $\displaystyle 2 a b + 2 a b$ $\displaystyle$ $=$ $\displaystyle 4 a b$

$\blacksquare$