Square of Difference/Algebraic Proof 2
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Theorem
- $\forall x, y \in \R: \paren {x - y}^2 = x^2 - 2 x y + y^2$
Proof
Follows directly from the Binomial Theorem:
- $\ds \forall n \in \Z_{\ge 0}: \paren {x + y}^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k$
putting $n = 2$ and $y = -y$.
$\blacksquare$