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.
 A: On OP's request, I am converting my comment into an answer.
If you want to put values of angles and sides, you can use 
$$a=\sqrt 3,\quad b=1,\quad c=2,\quad A=60^\circ,\quad B=30^\circ,\quad C=90^\circ$$
for which we have
$$\frac{c\sin(A-B)}{a^2-b^2}-\frac{b\sin(C-A)}{c^2-a^2}=\frac{2\sin(30^\circ)}{3-1}-\frac{\sin(30^\circ)}{4-3}=0$$
A: 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.
A: Assume that $a>b$.
There exists a point $D$ on line segment $AB$ such that $CD=b$, Produce $CD$ to meet the circumscribed circle of $\Delta ABC$ at point $E$. (See the figure below.)

Since $AC=CD$, $\angle CDA=\angle CAD=A$. Hence, $\angle BCD=A-B$.
Since $\angle BAE$ and $\angle BCE$ are angles in the same segment, $\angle BAE=\angle BCE=A-B$.
Applying the power theorem on the circle with center $C$ and radius $b$, we have $AB\cdot BD=a^2-b^2$. Since $AB=c$, $BD=\dfrac{a^2-b^2}{c}$.
Note that $\Delta BDE$ and $\Delta CDA$ are similar. We have $BE=BD=\dfrac{a^2-b^2}{c}$.
Applying the sine formula on $\Delta ABE$, we have $\dfrac{BE}{\sin\angle BAE}=2R$, where $R$ is the radius of the circumscribed circle of $\Delta ABC$. Hence, we have $\dfrac{c\sin(A-B)}{a^2-b^2}=\dfrac{1}{2R}$.
If $b>a$ instead, then we have $\dfrac{c\sin(B-A)}{b^2-a^2}=\dfrac{1}{2R}$. But $\dfrac{c\sin(A-B)}{a^2-b^2}$ and $\dfrac{c\sin(B-A)}{b^2-a^2}$ are indeed equal.
Similarly, we have $\dfrac{b\sin(C-A)}{c^2-a^2}=\dfrac{1}{2R}$. Thus, we have $\dfrac{c\sin(A-B)}{a^2-b^2}-\dfrac{b\sin(C-A)}{c^2-a^2}=0$.
A: 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
$$\frac{c\sin(A-B)}{a^2-b^2}-\frac{b\sin(C-A)}{c^2-a^2}=0$$
