In a triangle ABC, prove that ${sin}^{3} A cos (B - C) + {sin}^{3} B cos (C - A) + s i n^{3} C cos (A - B) = 3 sin A sin B sin C$ .

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Solution

${sin}^{3} A cos (B - C) = {sin}^{2} A sin A cos (B - C)$
$= \frac{1}{2} {sin}^{2} A \cdot 2 sin (B + C) cos (B - C)$
$= \frac{1}{2} {sin}^{2} A (sin 2 B + sin 2 C)$
$= {sin}^{2} A (sin B cos B + sin C cos C)$ .
$∴ L . H . S . = \sum {sin}^{3} A cos (B - C)$
$= {sin}^{2} A (sin B cos B + sin C cos C) + {sin}^{2} B (sin C cos C + sin A cos A) + {sin}^{2} C (sin A cos A + sin B cos B)$
$= sin A sin B (sin A cos B + cos A sin B) + . . . + . . .$
$= sin A sin B sin (A + B) + . . . + . . .$
$= sin A sin B sin C + . . . + . . .$
$= 3 sin A sin B sin C$ .

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