<s>[INST] Finding $\frac{c\sin(A-B)}{a^2-b^2}-\frac{b\sin(C-A)}{c^2-a^2}$ 
If $a, b$ and $c$ (all distinct) are the sides of a triangle ABC opposite to the angles $A, B$ and $C$, respectively, then $\frac{c\sin(A-B)}{a^2-b^2}-\frac{b\sin(C-A)}{c^2-a^2}$ is equal to $?$

By opening, $\sin(A-B)$ as $\sin A\cos B-\cos A\sin B$ and then $\sin A$ as $\frac{a}{2R}$ and $\cos A$ as $\frac{b^2+c^2-a^2}{2bc}$, I am able to get the answer as zero. But I am looking for a shorter solution, maybe by putting values of angles and sides. 
My first instinct was to assume the triangle to be equilateral. But the question invalidates that case. Then I thought of a right angled triangle with  pythogorean triplet as $3,4,5$. But here, I don't know other two angles. 
I wonder if I could just keep a dummy triangle handy whose sides and angles I know, which I could quickly use to solve such questions. Any help please? Thanks.
 [/INST] Use sine and cosine rules to evaluate  
\begin{align}
& \frac{\sin(A-B)}{\sin(A-C)}
= \frac{\sin A\cos B - \cos A \sin B}{\sin A\cos C - \cos A \sin C}\\
=& \frac{a\frac{a^2+c^2-b^2}{2ac} - b\frac{b^2+c^2-a^2}{2bc} }
{a\frac{a^2+b^2-c^2}{2ab} - c\frac{b^2+c^2-a^2}{2bc} }
 =\frac{\frac1c [(a^2+c^2-b^2 )- (b^2+c^2-a^2)] }
{\frac1b [(a^2+b^2-c^2 )- (b^2+c^2-a^2)] }
 =\frac{b(a^2-b^2)}
{c(a^2-c^2)}\\
\end{align}
Then, rearrange to obtain
$$\frac{c\sin(A-B)}{a^2-b^2}-\frac{b\sin(C-A)}{c^2-a^2}=0$$
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