<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] One can easily show that
$$
\frac{c\sin(A-B)}{a^2-b^2}=\frac1{2R},\tag1
$$
where $R$ is the radius of the circumscribed circle. 
Indeed substituting in LHS of (1)
$$a=2R\sin  A,\quad b=2R\sin  B,\quad c=2R\sin C $$
one obtains:
$$\begin{align}
\frac{c\sin(A-B)}{a^2-b^2}&=\frac1{2R}\frac{\sin C\sin(A-B)}{\sin^2A-\sin^2B}\\
&=\frac1{2R}\frac{\sin(A+B)\sin(A-B)}{\sin^2A-\sin^2B}\\
&=\frac1{2R}\frac{\frac12(\cos 2B-\cos 2A)}{\frac12(\cos 2B-\cos 2A)}.
\end{align}
$$
The same result one obtains of course for $\frac{b\sin(C-A)}{c^2-a^2}$ as well.
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