Difference of Squares of Sum and Difference/Algebraic Proof

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Theorem

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


Proof

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

$\blacksquare$