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In a $\triangle A B C, a, b, c$ are the sides of the triangle opposite to the angles $A, B, C,$ respectively. Then, the value of $a^{3} \sin (B-C)$ $+b^{3} \sin (C-A)+c^{3} \sin (A-B)$ is equal to
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$\sum a^{3} \sin (B-C)$
$=\sum k^{3} \sin ^{3} A \sin (B-C)$
$$
\begin{aligned}\left[\because \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k\right] \\=\sum k^{3}\left[\sin ^{2} A \sin (B+C) \sin (B-C)\right] \\=k^{3}\left[\left\{\sin ^{2} A \times \frac{1}{2}(\cos 2 C-\cos 2 B)\right\}\right.\\+\sin ^{2} B \times \frac{1}{2}(\cos 2 A-\cos 2 C) \\\left.+\sin ^{2} C \times \frac{1}{2}(\cos 2 B-\cos 2 A)\right] \end{aligned}
$$
$=\frac{k^{3}}{2}\left[\sin ^{2} A\left(1-2 \sin ^{2} C-1+2 \sin ^{2} B\right)\right.$
$+\sin ^{2} B\left(1-2 \sin ^{2} A-1+2 \sin ^{2} C\right)$
$\left.+\sin ^{2} C\left(1-2 \sin ^{2} B-1+2 \sin ^{2} A\right)\right]$
$=\frac{k^{3}}{2}\left[-2 \sin ^{2} A \sin ^{2} C+2 \sin ^{2} A \sin ^{2} B\right.$
$-2 \sin ^{2} B \sin ^{2} A+2 \sin ^{2} B \sin ^{2} C$
$\left.-2 \sin ^{2} C \sin ^{2} B+2 \sin ^{2} C \sin ^{2} A\right]$
$=\frac{k^{3}}{2}[0]=0$
$=\sum k^{3} \sin ^{3} A \sin (B-C)$
$$
\begin{aligned}\left[\because \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k\right] \\=\sum k^{3}\left[\sin ^{2} A \sin (B+C) \sin (B-C)\right] \\=k^{3}\left[\left\{\sin ^{2} A \times \frac{1}{2}(\cos 2 C-\cos 2 B)\right\}\right.\\+\sin ^{2} B \times \frac{1}{2}(\cos 2 A-\cos 2 C) \\\left.+\sin ^{2} C \times \frac{1}{2}(\cos 2 B-\cos 2 A)\right] \end{aligned}
$$
$=\frac{k^{3}}{2}\left[\sin ^{2} A\left(1-2 \sin ^{2} C-1+2 \sin ^{2} B\right)\right.$
$+\sin ^{2} B\left(1-2 \sin ^{2} A-1+2 \sin ^{2} C\right)$
$\left.+\sin ^{2} C\left(1-2 \sin ^{2} B-1+2 \sin ^{2} A\right)\right]$
$=\frac{k^{3}}{2}\left[-2 \sin ^{2} A \sin ^{2} C+2 \sin ^{2} A \sin ^{2} B\right.$
$-2 \sin ^{2} B \sin ^{2} A+2 \sin ^{2} B \sin ^{2} C$
$\left.-2 \sin ^{2} C \sin ^{2} B+2 \sin ^{2} C \sin ^{2} A\right]$
$=\frac{k^{3}}{2}[0]=0$
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