If $A + B + C = π$ , then prove that ${sin}^{3} A cos (B - C) + {sin}^{3} B cos (C - A) + {sin}^{3} C cos (A - B) = 3 sin A sin B sin C$

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Solution

LHS
$A + B + C = π A = - (B + C) + π {sin}^{3} A cos (B - C) + {sin}^{3} B cos (C - A) + {sin}^{3} C cos (A - B) {sin}^{2} A sin (B + C) cos (B - C) + {sin}^{2} B sin (A + C) cos (B - C) + {sin}^{2} C sin (A + B) cos (B - C) \frac{1}{2} {sin}^{2} A (sin 2 B + sin 2 C) + \frac{1}{2} {sin}^{2} B (sin 2 A + sin 2 C) + \frac{1}{2} {sin}^{2} C (sin 2 A + sin 2 B) \frac{1}{2} {sin}^{2} A (2 sin B cos B + 2 sin C cos C) + \frac{1}{2} {sin}^{2} B (2 sin C cos C + 2 sin A cos A) + \frac{1}{2} {sin}^{2} C (2 sin A cos A + 2 sin B cos B) sin A sin B (sin A cos B + cos A sin B) + sin B sin C (sin B cos C + cos B sin C) + sin A sin C (sin A cos C + cos A sin C) sin A sin B sin (A + B) + sin C sin B sin (B + C) + sin A sin C sin (A + C) 3 sin A sin B sin C$
LHS=RHS

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