In any $\triangle ABC$, prove that $\;$ $a^2 + b^2 + c^2 = 2 \left(b c \cos A + c a \cos B + a b \cos C\right)$
By cosine rule,
$\cos A = \dfrac{b^2 + c^2 - a^2}{2 b c}$, $\;$ $\cos B = \dfrac{c^2 + a^2 - b^2}{2 c a}$, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$
$\begin{aligned}
RHS & = 2 \left[b c \cos A + c a \cos B + a b \cos C\right] \\\\
& = 2 \left[b c \times \left(\dfrac{b^2 + c^2 - a^2}{2 b c}\right) + c a \times \left(\dfrac{c^2 + a^2 - b^2}{2 c a}\right) + a b \times \left(\dfrac{a^2 + b^2 - c^2}{2 a b}\right)\right] \\\\
& = b^2 + c^2 - a^2 + c^2 + a^2 - b^2 + a^2 + b^2 - c^2 \\\\
& = a^2 + b^2 + c^2 = LHS
\end{aligned}$
Hence proved.