In $\triangle ABC$, $\;$ prove that $\;$ $\dfrac{a \sin \left(B - C\right)}{b^2 - c^2} = \dfrac{b \sin \left(C - A\right)}{c^2 - a^2} = \dfrac{c \sin \left(A - B\right)}{a^2 - b^2}$
By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$
By cosine rule, $\;$ $\cos A = \dfrac{b^2 + c^2 - a^2}{2 b c}$, $\;$ $\cos B = \dfrac{c^2 + a^2 - b^2}{2 c a}$, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$
Now,
$\begin{aligned}
\sin \left(B - C\right) & = \sin B \cos C - \cos B \sin C \\\\
& = \dfrac{b}{2 R} \left(\dfrac{a^2 + b^2 - c^2}{2 a b}\right) - \dfrac{c}{2 R} \left(\dfrac{c^2 + a^2 - b^2}{2 c a}\right) \\\\
& = \dfrac{1}{4 a R} \left(a^2 + b^2 - c^2 - c^2 - a^2 + b^2\right) \\\\
& = \dfrac{2 \left(b^2 - c^2\right)}{4 a R} \\\\
& = \dfrac{b^2 - c^2}{2 a R} \;\;\; \cdots \; (1a)
\end{aligned}$
$\begin{aligned}
\sin \left(C - A\right) & = \sin C \cos A - \cos C \sin A \\\\
& = \dfrac{c}{2 R} \left(\dfrac{b^2 + c^2 - a^2}{2 b c}\right) - \dfrac{a}{2 R} \left(\dfrac{a^2 + b^2 - c^2}{2 a b}\right) \\\\
& = \dfrac{1}{4 b R} \left(b^2 + c^2 - a^2 - a^2 - b^2 + c^2\right) \\\\
& = \dfrac{2 \left(c^2 - a^2\right)}{4 b R} \\\\
& = \dfrac{c^2 - a^2}{2 b R} \;\;\; \cdots \; (1b)
\end{aligned}$
$\begin{aligned}
\sin \left(A - B\right) & = \sin A \cos B - \cos A \sin B \\\\
& = \dfrac{a}{2 R} \left(\dfrac{c^2 + a^2 - b^2}{2 c a}\right) - \dfrac{b}{2 R} \left(\dfrac{b^2 + c^2 - a^2}{2 b c}\right) \\\\
& = \dfrac{1}{4 c R} \left(c^2 + a^2 - b^2 - b^2 - c^2 + a^2\right) \\\\
& = \dfrac{2 \left(a^2 - b^2\right)}{4 c R} \\\\
& = \dfrac{a^2 - b^2}{2 c R} \;\;\; \cdots \; (1c)
\end{aligned}$
$\therefore \;$ In view of equation $(1a)$,
$\dfrac{a \sin \left(B - C\right)}{b^2 - c^2} = \left(\dfrac{a}{b^2 - c^2}\right) \times \left(\dfrac{b^2 - c^2}{2 a R}\right) = \dfrac{1}{2 R}$ $\;\;\; \cdots \; (2a)$
In view of equation $(1b)$,
$\dfrac{b \sin \left(C - A\right)}{c^2 - a^2} = \left(\dfrac{b}{c^2 - a^2}\right) \times \left(\dfrac{c^2 - a^2}{2 b R}\right) = \dfrac{1}{2 R}$ $\;\;\; \cdots \; (2b)$
In view of equation $(1c)$,
$\dfrac{c \sin \left(A - B\right)}{a^2 - b^2} = \left(\dfrac{c}{a^2 - b^2}\right) \times \left(\dfrac{a^2 - b^2}{2 c R}\right) = \dfrac{1}{2 R}$ $\;\;\; \cdots \; (2c)$
$\therefore \;$ From equations $(2a)$, $(2b)$ and $(2c)$ we have,
$\dfrac{a \sin \left(B - C\right)}{b^2 - c^2} = \dfrac{b \sin \left(C - A\right)}{c^2 - a^2} = \dfrac{c \sin \left(A - B\right)}{a^2 - b^2}$
Hence proved.