In $\triangle ABC$, prove that $a \sin \left(B - C\right) + b \sin \left(C - A\right) + c \sin \left(A - B\right) = 0$
By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$
$\therefore \;$ $a = 2 R \sin A$, $\;$ $b = 2 R \sin B$, $\;$ $c = 2 R \sin C$
$\begin{aligned}
\therefore \; a \sin \left(B - C\right) & = 2 R \sin A \left(\sin B \cos C - \cos B \sin C\right) \\\\
& = 2 R \left(\sin A \sin B \cos C - \sin A \cos B \sin C\right) \;\;\; \cdots \; (1a)
\end{aligned}$
$\begin{aligned}
b \sin \left(C - A\right) & = 2 R \sin B \left(\sin C \cos A - \cos C \sin A\right) \\\\
& = 2 R \left(\sin B \sin C \cos A - \sin B \cos C \sin A\right) \;\;\; \cdots \; (1b)
\end{aligned}$
$\begin{aligned}
c \sin \left(A - B\right) & = 2 R \sin C \left(\sin A \cos B - \cos A \sin B\right) \\\\
& = 2 R \left(\sin C \sin A \cos B - \sin C \cos A \sin B\right) \;\;\; \cdots \; (1c)
\end{aligned}$
Adding equations $(1a)$, $(1b)$ and $(1c)$ gives
$\begin{aligned}
LHS & = 2 R \left(\sin A \sin B \cos C - \sin A \cos B \sin C \right. \\\\
& \hspace{2cm} \left. + \sin B \sin C \cos A - \sin B \cos C \sin A \right. \\\\
& \hspace{3cm} \left. + \sin C \sin A \cos B - \sin C \cos A \sin B \right) \\\\
& = 0 = RHS
\end{aligned}$