In any $\triangle ABC$, prove that $\;$ $a \sin \left(\dfrac{A}{2}\right) \sin \left(\dfrac{B - C}{2}\right) + b \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C - A}{2}\right) + c \sin \left(\dfrac{C}{2}\right) \sin \left(\dfrac{A - B}{2}\right) = 0$
In $\triangle ABC$, $\;$ $A + B + C = \pi$
i.e. $\;$ $A = \pi - \left(B + C\right)$
i.e. $\;$ $\dfrac{A}{2} = \dfrac{\pi}{2} - \left(\dfrac{B + C}{2} \right)$
$\therefore \;$ $\sin \left(\dfrac{A}{2}\right) = \sin \left[\dfrac{\pi}{2} - \left(\dfrac{B + C}{2}\right)\right] = \cos \left(\dfrac{B + C}{2}\right)$
Similarly, $\;$ $\sin \left(\dfrac{B}{2}\right) = \cos \left(\dfrac{C + A}{2}\right)$
and $\;$ $\sin \left(\dfrac{C}{2}\right) = \cos \left(\dfrac{A + B}{2}\right)$
$\begin{aligned}
LHS & = a \sin \left(\dfrac{A}{2}\right) \sin \left(\dfrac{B - C}{2}\right) + b \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C - A}{2}\right) \\\\
& \hspace{5cm} + c \sin \left(\dfrac{C}{2}\right) \sin \left(\dfrac{A - B}{2}\right) \\\\
& = a \cos \left(\dfrac{B + C}{2}\right) \sin \left(\dfrac{B - C}{2}\right) + b \cos \left(\dfrac{C + A}{2}\right) \sin \left(\dfrac{C - A}{2}\right) \\\\
& \hspace{5.5cm} + c \cos \left(\dfrac{A + B}{2}\right) \sin \left(\dfrac{A - B}{2}\right) \\\\
& \left[\text{Note: } \sin A \cos B = \dfrac{1}{2} \left\{\sin \left(A + B\right) + \sin \left(A - B\right) \right\}\right] \\\\
& = \dfrac{a}{2} \left[\sin \left(\dfrac{B - C}{2} + \dfrac{B + C}{2}\right) + \sin \left(\dfrac{B - C}{2} - \dfrac{B + C}{2}\right)\right] \\\\
& \hspace{1cm} + \dfrac{b}{2} \left[\sin \left(\dfrac{C - A}{2} + \dfrac{C + A}{2}\right) + \sin \left(\dfrac{C - A}{2} - \dfrac{C + A}{2}\right)\right] \\\\
& \hspace{2cm} + \dfrac{c}{2} \left[\sin \left(\dfrac{A - B}{2} + \dfrac{A + B}{2}\right) + \sin \left(\dfrac{A - B}{2} - \dfrac{A + B}{2}\right)\right] \\\\
& = \dfrac{a}{2} \left[\sin B + \sin \left(- C\right)\right] + \dfrac{b}{2} \left[\sin C + \sin \left(- A\right)\right] + \dfrac{c}{2} \left[\sin A + \sin \left(- B\right)\right] \\\\
& = \dfrac{a}{2} \left[\sin B - \sin C\right] + \dfrac{b}{2} \left[\sin C - \sin A\right] + \dfrac{c}{2} \left[\sin A - \sin B\right] \\\\
& = \dfrac{a \sin B}{2} - \dfrac{a \sin C}{2} + \dfrac{b \sin C}{2} - \dfrac{b \sin A}{2} + \dfrac{c \sin A}{2} - \dfrac{c \sin B}{2} \\\\
& = \dfrac{\sin A}{2} \left(c - b\right) + \dfrac{\sin B}{2} \left(a - c\right) + \dfrac{\sin C}{2} \left(b - a\right) \\\\
& \left[\text{Note: Sine rule: } \dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2 R \right. \\
& \hspace{2.75cm} \left. R = \text{circumradius of } \triangle ABC \right. \\
& \hspace{2.75cm} \left. \implies \sin A = \dfrac{a}{2 R}, \; \sin B = \dfrac{b}{2 R}, \; \sin C = \dfrac{c}{2 R} \right] \\\\
& = \dfrac{a}{4 R} \left(c - b\right) + \dfrac{b}{4 R} \left(a - c\right) + \dfrac{c}{4 R} \left(b - a\right) \\\\
& = \dfrac{1}{4 R} \left[a c - a b + a b - b c + b c - a c\right] \\\\
& = 0 = RHS
\end{aligned}$
Hence proved.