[Geometry]
Problem: In triangle $A B C, a=12, b=17$, and $c=13$. Compute $b \cos C-c \cos B$.

Solution: Using the Law of Cosines, $a^{2}+b^{2}-2 a b \cos C=c^{2}$ implies

$$
b \cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a}
$$

Similarly,

$$
c \cos B=\frac{a^{2}-b^{2}+c^{2}}{2 a}
$$

Thus

$$
\begin{aligned}
b \cos C-c \cos B & =\frac{a^{2}+b^{2}-c^{2}}{2 a}-\frac{a^{2}-b^{2}+c^{2}}{2 a} \\
& =\frac{2 b^{2}-2 c^{2}}{2 a} \\
& =\frac{b^{2}-c^{2}}{a} .
\end{aligned}
$$



With the given values, the result is $\left(17^{2}-13^{2}\right) / 12=120 / 12=\mathbf{1 0}$.
Let $H$ be the foot of the altitude from $A$ to $\overline{B C}$; let $B H=x$, $C H=y$, and $A H=h$. Then $b \cos C=y, c \cos B=x$, and the desired quantity is $Q=y-x$. However, $y+x=a$, so $y^{2}-x^{2}=a Q$. By the Pythagorean Theorem, $y^{2}=b^{2}-h^{2}$ and $x^{2}=c^{2}-h^{2}$, so $y^{2}-x^{2}=\left(b^{2}-h^{2}\right)-\left(c^{2}-h^{2}\right)=b^{2}-c^{2}$. Thus $a Q=b^{2}-c^{2}$, and $Q=\frac{b^{2}-c^{2}}{a}$

With the given values, the result is $\left(17^{2}-13^{2}\right) / 12=120 / 12=\mathbf{1 0}$.

Final answer: 10