In any $\triangle ABC$, prove that $\;$ $a^3 \cos \left(B - C\right) + b^3 \cos \left(C - A\right) + c^3 \cos \left(A - B\right) = 3 a b c$
By cosine rule, $\;$ $\cos B = \dfrac{c^2 + a^2 - b^2}{2 c a}$, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$
By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$
$\implies$ $a = 2 R \sin A$, $\;$ $b = 2 R \sin B$, $\;$ $c = 2 R \sin C$
In $\triangle ABC$, $\;$ $A + B + C = \pi$
$\begin{aligned}
LHS & = a^3 \cos \left(B - C\right) + b^3 \cos \left(C - A\right) + c^3 \cos \left(A - B\right) \\\\
& = a^2 \times a \cos \left(B - C\right) + b^2 \times b \cos \left(C - A\right) + c^2 \times c \cos \left(A - B\right) \\\\
& = 2 R \left[a^2 \sin A \cos \left(B - C\right) + b^2 \sin B \cos \left(C - A\right) \right. \\
& \hspace{4.5cm} \left. + c^2 \sin C \cos \left(A - B\right)\right] \;\; \left[\text{By sine rule}\right] \\\\
& \left[\text{Note: In } \triangle ABC, \; A + B + C = \pi \right. \\
& \hspace{2.5cm} \left. i.e. \; A = \pi - \left(B + C\right) \right. \\
& \hspace{2.5cm} \left. \therefore \; \sin A = \sin \left[\pi - \left(B + C\right)\right] = \sin \left(B + C\right) \right. \\
& \left. \text{Similarly, } \sin B = \sin \left(C + A\right), \; \sin C = \sin \left(A + B\right) \right] \\\\
& = 2 R \left[a^2 \sin \left(B + C\right) \cos \left(B - C\right) \right. \\
& \hspace{2cm} \left. + b^2 \sin \left(C + A\right) \cos \left(C - A\right) \right. \\
& \hspace{4cm} \left. + c^2 \sin \left(A + B\right) \cos \left(A - B\right) \right] \\\\
& \left[\text{Note: } \sin \alpha \cos \beta = \dfrac{1}{2} \left[\sin \left(\alpha + \beta\right) + \sin \left(\alpha - \beta\right)\right]\right] \\\\
& = 2 R \left\{\dfrac{a^2}{2} \left[\sin \left(B + C + B - C\right) + \sin \left(B + C - B + C\right)\right] \right. \\
& \hspace{2cm} \left. + \dfrac{b^2}{2} \left[\sin \left(C + A + C - A\right) + \sin \left(C + A - C + A\right)\right] \right. \\
& \hspace{3cm} \left. + \dfrac{c^2}{2} \left[\sin \left(A + B + A - B\right) + \sin \left(A + B - A + B\right)\right] \right\} \\\\
& = 2 R \left\{\dfrac{a^2}{2} \left(\sin 2 B + \sin 2 C\right) + \dfrac{b^2}{2} \left(\sin 2 C + \sin 2 A\right) + \dfrac{c^2}{2} \left(\sin 2 A + \sin 2 B\right) \right\} \\\\
& = R \left(b^2 + c^2\right) \sin 2 A + R \left(c^2 + a^2\right) \sin 2 B + R \left(a^2 + b^2\right) \sin 2 C
\end{aligned}$
$\begin{aligned}
\left[\text{Note: } R \left(a^2 + b^2\right) \sin 2 C \right. & \left. = R \left(a^2 + b^2\right) \times 2 \sin C \cos C \right. \\
\left. \right. & \left. = R \left(a^2 + b^2\right) \times 2 \times \dfrac{c}{2 R} \times \left(\dfrac{a^2 + b^2 - c^2}{2 a b}\right) \right. \\
\left. \right. & \left. = \dfrac{c}{2 a b} \left(a^2 + b^2\right) \left(a^2 + b^2 - c^2\right) \right. \\
\left. \right. & \left. = \dfrac{c^2}{2 a b c} \left(a^2 + b^2\right) \left(a^2 + b^2 - c^2\right) \right. \\
\left. \text{Similarly, } \; R \left(b^2 + c^2\right) \sin 2 A \right. & \left. = \dfrac{a^2}{2 a b c} \left(b^2 + c^2\right) \left(b^2 + c^2 - a^2\right) \right. \\
\left. \text{and, } \; R \left(c^2 + a^2\right) \sin 2 B \right. & \left. = \dfrac{b^2}{2 a b c} \left(c^2 + a^2\right) \left(c^2 + a^2 - b^2\right) \right]
\end{aligned}$
$\begin{aligned}
\therefore \; LHS & = \dfrac{1}{2 a b c} \left[a^2 \left(b^2 + c^2\right) \left(b^2 + c^2 - a^2\right) \right. \\
& \hspace{2cm} \left. + b^2 \left(c^2 + a^2\right) \left(c^2 + a^2 - b^2\right) \right. \\
& \hspace{3cm} \left. + c^2 \left(a^2 + b^2\right) \left(a^2 + b^2 - c^2\right) \right] \\\\
& = \dfrac{1}{2 a b c} \left[a^2 \left(b^4 + c^4 + 2 b^2 c^2 - a^2 b^2 - a^2 c^2\right) \right. \\
& \hspace{2cm} \left. + b^2 \left(c^4 + a^4 + 2 c^2 a^2 - b^2 c^2 - a^2 b^2\right) \right. \\
& \hspace{3cm} \left. + c^2 \left(a^4 + b^4 + 2 a^2 b^2 - a^2 c^2 - b^2 c^2\right) \right] \\\\
& = \dfrac{1}{2 a b c} \left[a^2 b^4 + a^2 c^4 + 2 a^2 b^2 c^2 - a^4 b^2 - a^4 c^2 \right. \\
& \hspace{2cm} \left. + b^2 c^4 + a^4 b^2 + 2 a^2 b^2 c^2 - b^4 c^2 - a^2 b^4 \right. \\
& \hspace{3cm} \left. + a^4 c^2 + b^4 c^2 + 2 a^2 b^2 c^2 - a^2 c^4 - b^2 c^4 \right] \\\\
& = \dfrac{6 a^2 b^2 c^2}{2 a b c} \\\\
& = 3 a b c = RHS
\end{aligned}$
Hence proved.