In $\triangle ABC$, prove that $\left(b^2 - c^2\right) \cot A + \left(c^2 - a^2\right) \cot B + \left(a^2 - b^2\right) \cot C = 0$
By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$
$\implies$ $a = 2 R \sin A$, $\;$ $b = 2 R \sin B$, $\;$ $c = 2 R \sin C$
Now,
$\begin{aligned}
\left(b^2 - c^2\right) \cot A & = \left(4 R^2 \sin^2 B - 4 R^2 \sin^2 C\right) \times \dfrac{\cos A}{\sin A} \;\;\; \left[\text{by sine rule}\right] \\\\
& = 4 R^2 \left(\sin^2 B - \sin^2 C\right) \times \dfrac{\cos A}{\sin A} \\\\
& \left[\text{Note: } \sin^2 \alpha - \sin^2 \beta = \sin \left(\alpha + \beta\right) \sin \left(\alpha - \beta \right)\right] \\\\
& = 4 R^2 \sin \left(B + C\right) \sin \left(B - C\right) \times \dfrac{\cos A}{\sin A} \\\\
& \left[\text{In } \triangle ABC, \; A + B + C = \pi \right. \\\\
& \left. \therefore \; B + C = \pi - A \right. \\\\
& \left. \therefore \; \sin \left(B + C\right) = \sin \left(\pi - A\right) = \sin A \right] \\\\
& = 4 R^2 \sin A \sin \left(B - C\right) \times \dfrac{\cos A}{\sin A} \\\\
& = 4 R^2 \sin \left(B - C\right) \cos A \\\\
& = 4 R^2 \left(\sin B \cos C - \cos B \sin C\right) \cos A \\\\
& = 4 R^2 \left(\cos A \sin B \cos C - \cos A \cos B \sin C\right) \;\;\; \cdots \; (1a)
\end{aligned}$
Similarly,
$\begin{aligned}
\left(c^2 - a^2\right) \cot B & = \left(4 R^2 \sin^2 C - 4 R^2 \sin^2 A\right) \times \dfrac{\cos B}{\sin B} \\\\
& = 4 R^2 \left(\cos A \cos B \sin C - \sin A \cos B \cos C\right) \;\;\; \cdots \; (1b)
\end{aligned}$
$\begin{aligned}
\left(a^2 - b^2\right) \cot C & = \left(4 R^2 \sin^2 A - 4 R^2 \sin^2 B\right) \times \dfrac{\cos C}{\sin C} \\\\
& = 4 R^2 \left(\sin A \cos B \cos C - \cos A \sin B \cos C\right) \;\;\; \cdots \; (1c)
\end{aligned}$
Adding equations $(1a)$, $(1b)$ and $(1c)$ we have,
$\begin{aligned}
LHS & = \left(b^2 - c^2\right) \cot A + \left(c^2 - a^2\right) \cot B + \left(a^2 - b^2\right) \cot C \\\\
& = 4 R^2 \left[ \cos A \sin B \cos C - \cos A \cos B \sin C \right. \\\\
& \left. \hspace{1.5cm} + \cos A \cos B \sin C - \sin A \cos B \cos C \right. \\\\
& \left. \hspace{2cm} + \sin A \cos B \cos C - \cos A \sin B \cos C \right] \\\\
& = 0 = RHS
\end{aligned}$
Hence proved.