In $\triangle ABC$, prove that $a^2 \left(\cos^2 B - \cos^2 C\right) + b^2 \left(\cos^2 C - \cos^2 A\right) + c^2 \left(\cos^2 A - \cos^2 B\right) = 0$
By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$
$\implies$ $\sin A = \dfrac{a}{2 R}$, $\;$ $\sin B = \dfrac{b}{2 R}$, $\;$ $\sin C = \dfrac{c}{2 R}$
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, $\;$ $\cos^2 B - \cos^2 C $
$ = \left(\cos B + \cos C\right) \left(\cos B - \cos C\right) $
$ = 2 \cos \left(\dfrac{B + C}{2}\right) \cos \left(\dfrac{B - C}{2}\right) \times \left(-2\right) \sin \left(\dfrac{B + C}{2}\right) \sin \left(\dfrac{B - C}{2}\right) $
In $\triangle ABC$, $\;$ $B + C = \pi - A$
i.e. $\;$ $\dfrac{B + C}{2} = \dfrac{\pi}{2} - \dfrac{A}{2}$
$\therefore \;$ $\cos \left(\dfrac{B + C}{2}\right) = \cos \left(\dfrac{\pi}{2} - \dfrac{A}{2}\right) = \sin \left(\dfrac{A}{2}\right)$
and, $\;$ $\sin \left(\dfrac{B + C}{2}\right) = \sin \left(\dfrac{\pi}{2} - \dfrac{A}{2}\right) = \cos \left(\dfrac{A}{2}\right)$
$\therefore \;$ $\cos^2 B - \cos^2 C$
$= - 2 \sin \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B - C}{2}\right) \times 2 \cos \left(\dfrac{A}{2}\right) \sin \left(\dfrac{B - C}{2}\right)$
$= - \sin A \sin \left(B - C\right)$ $\;\;\; \cdots \; (1a)$
Similarly, $\;$ $\cos^2 C - \cos^2 A = - \sin B \sin \left(C - A\right)$ $\;\;\; \cdots \; (1b)$
and $\;$ $\cos^2 A - \cos^2 B = - \sin C \sin \left(A - B\right)$ $\;\;\; \cdots \; (1c)$
$\therefore \;$ In view of equations $(1a)$, $(1b)$ and $(1c)$ we have,
$LHS = a^2 \left(\cos^2 B - \cos^2 C\right) + b^2 \left(\cos^2 C - \cos^2 A\right) + c^2 \left(\cos^2 A - \cos^2 B\right)$
$= - a^2 \sin A \sin \left(B - C\right) - b^2 \sin B \sin \left(C - A\right) - c^2 \sin C \sin \left(A - B\right)$ $\;\;\; \cdots \; (2)$
$\therefore \;$ Using sine rule, equation $(2)$ can be written as
$LHS = - \dfrac{1}{2 R} \left\{ a^3 \sin \left(B - C\right) + b^3 \sin \left(C - A\right) + c^3 \sin \left(A - B\right) \right\}$
$= - \dfrac{1}{2 R} \left\{a^3 \left(\sin B \cos C - \cos B \sin C\right) \right.$
$\left. \hspace{2cm} + b^3 \left(\sin C \cos A - \cos C \sin A\right) \right.$
$\left. \hspace{3cm} + c^3 \left(\sin A \cos B - \cos A \sin B\right) \right\}$ $\;\;\; \cdots \; (3)$
$\therefore \;$ Using sine rule and cosine rule, equation $(3)$ can be written as
$LHS = \dfrac{-1}{2R} \left\{a^3 \left[\left(\dfrac{b}{2R}\right) \times \left(\dfrac{a^2 + b^2 - c^2}{2 a b}\right) - \left(\dfrac{c}{2R}\right) \times \left(\dfrac{c^2 + a^2 - b^2}{2 c a}\right)\right] \right.$
$\left. \hspace{2cm} + b^3 \left[\left(\dfrac{c}{2R}\right) \times \left(\dfrac{b^2 + c^2 - a^2}{2 b c}\right) - \left(\dfrac{a}{2R}\right) \times \left(\dfrac{a^2 + b^2 - c^2}{2 a b}\right)\right] \right.$
$\left. \hspace{3cm} + c^3 \left[\left(\dfrac{a}{2R}\right) \times \left(\dfrac{c^2 + a^2 - b^2}{2 c a}\right) - \left(\dfrac{b}{2R}\right) \times \left(\dfrac{b^2 + c^2 - a^2}{2 b c}\right)\right] \right\}$
$= \dfrac{-1}{8R^2} \left\{a^2 \left[\left(a^2 + b^2 - c^2\right) - \left(c^2 + a^2 - b^2\right)\right] \right.$
$\left. \hspace{2.5cm} + b^2 \left[\left(b^2 + c^2 - a^2\right) - \left(a^2 + b^2 - c^2\right)\right] \right.$
$\left. \hspace{3.5cm} + c^2 \left[\left(c^2 + a^2 - b^2\right) - \left(b^2 + c^2 - a^2\right)\right] \right\}$
$= \dfrac{-1}{8R^2} \left\{a^2 \left(2 b^2 - 2 c^2\right) + b^2 \left(2 c^2 - 2 a^2\right) + c^2 \left(2 a^2 - b^2\right) \right\}$
$= \dfrac{-1}{4 R^2} \left\{a^2 b^2 - a^2 c^2 + b^2 c^2 - b^2 a^2 + c^2 a^2 - c^2 b^2 \right\}$
$= 0 = RHS$
Hence proved.