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In any $\triangle A B C, a(b \cos C-c \cos B)$ equals
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The correct answer is:
$b^2-c^2$
$\begin{aligned} a(b \cos C-c \cos B) & =a b \cos C-a c \cos B \\ & =\frac{a^2+b^2-c^2}{2}-\frac{a^2+c^2-b^2}{2} \\ & =b^2-c^2\end{aligned}$
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