Proving the following trigonometric proportion $$\frac{a\sin(B-C)}{b^2-c^2}=\frac{b\sin(C-A)}{c^2-a^2}=\frac{c\sin(A-B)}{a^2-b^2}$$
A, B, C are angles of a triangle and a, b, c are the sides of a triangle.
I tried using various things such as sine rule and then replacing the various rations in terms of sides of triangle, however it produced no result. Urgent help is needed!

By the Law of Sines, we have
$$a = d \sin A \qquad b = d \sin B \qquad c = d \sin C$$
where $d$ is the diameter of the triangle's circumcircle. As a result, the relation to prove becomes (after eliminating some $d$s)
$$\frac{\sin A \;\sin(B-C)}{\sin^2B-\sin^2C} = \frac{\sin B \;\sin(C-A)}{\sin^2C-\sin^2A} = \frac{\sin C \;\sin(A-B)}{\sin^2A-\sin^2B} \qquad (\star)$$
Consider $\sin A\;\sin(B-C)$, where we know $A = \pi - (B+C)$ so that $\sin A = \sin(B+C)$. By a Product-to-Sum Identity,
$$\begin{align}
\sin(B+C) \sin(B-C) &= \frac{1}{2}\left(\;\cos((B+C)-(B-C))- \cos((B+C)+(B-C)) \;\right) \\[4pt]
&= \frac{1}{2}\left( \cos 2C - \cos 2B \right) \\[4pt]
&= \frac{1}{2}\left( \; ( 1 - 2 \sin^2 C ) - ( 1 - 2 \sin^2 B )\; \right) \\[4pt]
&= \sin^2 B - \sin^2 C
\end{align}$$
Thus, the left-most expression of $(\star)$ reduces to $1$; by symmetry, the other sides do, as well, so that all three are in fact equal. (Note that the expressions in the original problem reduce to $1/d$.)
