Prove that: $a^{3} cos B cos C + b^{3} cos C cos A + c^{3} cos A cos B = a b c (1 - 2 cos A cos B cos C) .$

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Solution

L.H.S.= $k^{3} [{sin}^{3} A cos B cos C + {sin}^{3} B cos C cos A + {sin}^{3} C cos A cos B]$
= $k^{3} [{sin}^{3} A cos B cos C + sin B (1 - {cos}^{2} B) cos C cos A + sin C (1 - {cos}^{2} C) cos A cos B]$
$k^{3} [{sin}^{3} A cos B cos C + cos A (sin B cos C + sin C cos B) - \frac{1}{2} cos A cos B cos C (sin 2 B + sin 2 C)]$
$= k^{3} [{sin}^{3} A cos B cos C + cos A sin (B + C) - cos A cos B cos C sin (B + C) cos (B - C)]$
$= k^{3} [sin A (1 - {cos}^{2} A) cos B cos C + cos A sin A - cos A cos B cos C sin A cos (B - C)]$
= $k^{3} sin A [cos B cos C (1 - {cos}^{2} A) + cos A - cos A cos B cos C cos (B - C)]$
= $k^{3} sin A [cos A + cos B cos C - cos A cos B cos C cos A + cos (B - C)]$
= $k^{3} sin A [- cos (B + C) + cos B cos C - cos A cos B cos C - cos (B + C) + cos (B - C)]$
= $k^{3} sin A [sin A sin C - cos A cos B cos C . 2 sin B sin C]$
$= k^{3} sin A sin B sin C (1 - 2 cos A cos B cos C)$
= $a b c (1 - 2 cos A cos B cos C) .$ = R.H.S

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