Sum of Squares of Sum and Difference/Algebraic Proof

Theorem

$\forall a, b \in \R: \paren {a + b}^2 + \paren {a - b}^2 = 2 \paren {a^2 + b^2}$

Proof

\(\ds \)

\(\)

\(\ds \left({a + b}\right)^2 + \left({a - b}\right)^2\)

\(\ds \)

\(=\)

\(\ds a^2 + 2 a b + b^2 + a^2 - 2 a b + b^2\)

Square of Sum and Square of Difference

\(\ds \)

\(=\)

\(\ds a^2 + b^2 + a^2 + b^2\)

\(\ds \)

\(=\)

\(\ds 2 \left({a^2 + b^2}\right)\)

$\blacksquare$