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In any $\triangle \mathrm{ABC}$, with usual notations, $\mathrm{c}(\mathrm{a} \cos \mathrm{B}-\mathrm{b} \cos \mathrm{A})=$
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The correct answer is:
$a^2-b^2$
$\begin{aligned} & c(a \cos B-b \cos A) \\ & =a c\left(\frac{c^2+a^2-b^2}{2 a c}\right)-b c\left(\frac{b^2+c^2-a^2}{2 b c}\right) \\ & =\frac{c^2+a^2-b^2}{2}-\frac{b^2+c^2-a^2}{2}=\frac{2 a^2-2 b^2}{2}=a^2-b^2\end{aligned}$
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