Prove $sin 3 A {cos}^{3} (B - C) + sin 3 B {cos}^{3} (C - A) + sin 3 C {cos}^{3} (A - B) = sin 3 A sin 3 B sin 3 C$ .

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Solution

$cos 3 θ = 4 {cos}^{3} θ = 3 cos θ,$
$∴ c o s^{3} θ = \frac{1}{4} (cos 3 θ + 3 cos θ)$ .
$∴ sin 3 A {cos}^{3} (B - C)$
$= \frac{1}{4} sin 3 A [cos 3 (B - C) + 3 cos (B - C)]$
$= \frac{1}{4} sin 3 (B + C) cos 3 (B - C) + \frac{3}{4} sin (3 B + 3 C) cos (B - C)$
$= \frac{1}{8} (sin 6 B + sin 6 C) + \frac{3}{8} {sin (4 B + 2 C) + sin (2 B + 4 C)}$
Now $4 B + 2 C = 2 (B + C) + 2 B = 2 π - 2 A + 2 B$
$sin (4 B + 2 C) = sin {2 π + 2 (B - A)} = sin 2 (B - A)$ .
$∴$ L.H.S. $= \frac{1}{8} \sum (sin 6 B + sin 6 C) + \frac{3}{8} \sum [sin 2 (B - A) + sin 2 (C - B)]$
$∴ \sum sin 3 A {cos}^{3} (B - C) = \frac{1}{8} \cdot 2 \sum sin 6 A + \frac{3}{8} \cdot 0$
$= \frac{1}{4} (sin 6 A + sin 6 B + sin 6 C)$
$= \frac{1}{4} \cdot 4 sin 3 A sin 3 B sin 3 C$ ,
$= sin 3 A sin 3 B sin 3 C$ .

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