Source: Regional Olympiad - Federation of Bosnia and Herzegovina 2018
Tags: algebra, identity, real numbers
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if $a$, $b$ and $c$ are real numbers such that $(a-b)(b-c)(c-a) \neq 0$, prove the equality:
$\frac{b^2c^2}{(a-b)(a-c)}+\frac{c^2a^2}{(b-c)(b-a)}+\frac{a^2b^2}{(c-a)(c-b)}=ab+bc+ca$
Just an algebra bash
Solution
\begin{align*}
\frac{b^2c^2}{(a-b)(a-c)}+\frac{c^2a^2}{(b-c)(b-a)}+\frac{a^2b^2}{(c-a)(c-b)} &= -\frac{b^2c^2(b-c)}{(a-b)(b-c)(c-a)}-\frac{c^2a^2(c-a)}{(a-b)(b-c)(c-a)}-\frac{a^2b^2(a-b)}{(a-b)(b-c)(c-a)}\\
&= -\frac{b^2c^2(b-c)+c^2a^2(c-a)+a^2b^2(a-b)}{(a-b)(b-c)(c-a)}\\
&= \frac{b^2c^3-b^3c^2+c^2a^3-c^3a^2+a^2b^3-a^3b^2}{(a-b)(b-c)(c-a)}\\
&= \frac{(ab+bc+ca)(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}\\
&= ab+bc+ca
\end{align*}