Prove that $a^{3} sin (B - C) + b^{3} sin (C - A) + c^{3} sin (A - B) = 0.$

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Solution

$a^{2} - sin (B - C) = k^{2} {sin}^{3} A sin (B - C)$
$= k^{3} - {sin}^{2} A sin (B + C) = sin (B - C)$
$[∵ sin A = sin (B + C)]$
$= k^{3} - {sin}^{2} A ({sin}^{2} B - {sin}^{2} C)$
Hence $L . H . S = k^{3} [{sin}^{2} A ({sin}^{2} B - {sin}^{2} C) + {sin}^{2} B ({sin}^{2} C - {sin}^{2} A) + {sin}^{2} C ({sin}^{2} A - {sin}^{2} B)] = 0$ Hence proved

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