Example 6 In $\triangle A B C$, find the value of $a^{3} \sin (B-C)+b^{3} \sin (C-A)+c^{3} \sin (A-B)$.
First, we prove a lemma:
Lemma: In $\triangle ABC$, we have
$$
\frac{a^{2}-b^{2}}{c^{2}}=\frac{\sin (A-B)}{\sin C}, \frac{c^{2}-a^{2}}{b^{2}}=\frac{\sin (C-A)}{\sin B}, \frac{b^{2}-c^{2}}{a^{2}}=\frac{\sin (B-C)}{\sin A}.
$$

In fact, it is sufficient to prove one of them, let's prove $\frac{a^{2}-b^{2}}{c^{2}}=\frac{\sin (A-B)}{\sin C}$.
$$
\begin{array}{l}
\frac{a^{2}-b^{2}}{c^{2}}=\frac{\sin ^{2} A-\sin ^{2} B}{\sin ^{2} C}=\frac{(\sin A-\sin B)(\sin A+\sin B)}{\sin ^{2} C} \\
= \frac{2 \sin \frac{A+B}{2} \cos \frac{A-B}{2} \cdot 2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}}{\sin (A+B) \sin C}=\frac{\sin (A+B) \sin (A-B)}{\sin (A+B) \sin C}=\frac{\sin (A-B)}{\sin C}.
\end{array}
$$

Now, returning to the original problem, since $\frac{a^{2}-b^{2}}{c^{2}}=\frac{\sin (A-B)}{\sin C}$,
we have $c^{3} \sin (A-B)=c\left(a^{2}-b^{2}\right) \sin C=\frac{c^{2}\left(a^{2}-b^{2}\right)}{2 R}$.
Similarly, $b^{3} \sin (C-A)=\frac{b^{2}\left(c^{2}-a^{2}\right)}{2 R}, a^{3} \sin (B-C)=\frac{a^{2}\left(b^{2}-c^{2}\right)}{2 R}$.
Adding these three equations, we get
$$
a^{3} \sin (B-C)+b^{3} \sin (C-A)+c^{3} \sin (A-B)=\frac{c^{2}\left(a^{2}-b^{2}\right)+b^{2}\left(c^{2}-a^{2}\right)+a^{2}\left(b^{2}-c^{2}\right)}{2 R}=0,
$$

Therefore, the value of the required expression is 0.