<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] 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 $\dfrac{c\sin(A-B)}{a^2-b^2}-\dfrac{b\sin(C-A)}{c^2-a^2}=0$.
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