Q.In $\triangle ABC$, prove that $a^3\sin(B-C) + b^3\sin(C-A) + c^3\sin(A-B) = 0$.

By the companion identity $a^2\sin(B-C)=(b^2-c^2)\sin A$ (proved the same way as Miscellaneous Exercise 3, Q11C, using $\sin^2X-\sin^2Y=\sin(X+Y)\sin(X-Y)$ and $B+C=\pi-A$): $a^3\sin(B-C)=a\cdot a^2\sin(B-C)=a(b^2-c^2)\sin A=(a\sin A)(b^2-c^2)$. By the Sine Rule, $\sin A=\dfrac{a}{2R}$, so $a\sin A=\dfrac{a^2}{2R}$, giving $a^3\sin(B-C)=\dfrac{a^2(b^2-c^2)}{2R}$. Cyclically, $b^3\sin(C-A)=\dfrac{b^2(c^2-a^2)}{2R}$ and $c^3\sin(A-B)=\dfrac{c^2(a^2-b^2)}{2R}$. Adding all three: $a^3\sin(B-C)+b^3\sin(C-A)+c^3\sin(A-B)=\dfrac{1}{2R}\left[a^2(b^2-c^2)+b^2(c^2-a^2)+c^2(a^2-b^2)\right]$. The …

Factor out one power of $a$ (resp. $b,c$) to reuse the simpler identity $a^2\sin(B-C)=(b^2-c^2)\sin A$, apply the Sine Rule once more to reach a purely algebraic cyclic sum, then …

Attempting a direct triple-angle expansion of $\sin^3A\sin(B-C)$ from scratch, which is far longer and more error-pron …