In $\triangle ABC$, prove that $\left(\dfrac{b^2 - c^2}{a^2}\right) \sin 2A + \left(\dfrac{c^2 - a^2}{b^2}\right) \sin 2B + \left(\dfrac{a^2 - b^2}{c^2}\right) \sin 2 C = 0$
By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$
$\implies$ $\sin A = \dfrac{a}{2 R}$, $\;$ $\sin B = \dfrac{b}{2 R}$, $\;$ $\sin C = \dfrac{c}{2 R}$
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,
$\begin{aligned}
\left(\dfrac{b^2 - c^2}{a^2}\right) \sin 2 A & = \left(\dfrac{b^2 - c^2}{a^2}\right) \times 2 \sin A \cos A \\\\
& = \left(\dfrac{b^2 - c^2}{a^2}\right) \times 2 \times \dfrac{a}{2 R} \times \left(\dfrac{b^2 + c^2 - a^2}{2 b c}\right) \\
& \hspace{3cm} \left[\text{by sine and cosine rules}\right] \\\\
& = \dfrac{\left(b^2 - c^2\right) \left(b^2 + c^2 - a^2\right)}{2 R a b c} \\\\
& = \dfrac{b^4 + b^2 c^2 - b^2 a^2 - b^2 c^2 - c^4 + a^2 c^2}{2 R a b c} \\\\
& = \dfrac{b^4 - b^2 a^2 - c^4 + a^2 c^2}{2 R a b c} \;\;\; \cdots \; (1a)
\end{aligned}$
Similarly,
$\begin{aligned}
\left(\dfrac{c^2 - a^2}{b^2}\right) \sin 2 B & = \left(\dfrac{c^2 - a^2}{b^2}\right) \times 2 \sin B \cos B \\\\
& = \dfrac{c^4 - b^2 c^2 -a^4 + a^2 b^2}{2 R a b c} \;\;\; \cdots \; (1b)
\end{aligned}$
$\begin{aligned}
\left(\dfrac{a^2 - b^2}{c^2}\right) \sin 2 C & = \left(\dfrac{a^2 - b^2}{c^2}\right) \times 2 \sin C \cos C \\\\
& = \dfrac{a^4 - a^2 c^2 - b^4 + b^2 c^2}{2 R a b c} \;\;\; \cdots \; (1c)
\end{aligned}$
Adding equations $(1a)$, $(1b)$ and $(1c)$ we have,
$\begin{aligned}
LHS & = \left(\dfrac{b^2 - c^2}{a^2}\right) \sin 2A + \left(\dfrac{c^2 - a^2}{b^2}\right) \sin 2B + \left(\dfrac{a^2 - b^2}{c^2}\right) \sin 2 C \\\\
& = \dfrac{1}{2 R a b c} \left(b^4 - a^2 b^2 - c^4 + a^2 c^2 \right. \\\\
& \left. \hspace{2.5cm} + c^4 - b^2 c^2 - a^4 + a^2 b^2 \right. \\\\
& \left. \hspace{3.5cm} + a^4 - a^2 c^2 - b^4 + b^2 c^2 \right) \\\\
& = 0 = RHS
\end{aligned}$
Hence proved.