Is $cos (A + B + C) = cos A cos B cos C$ , then
$8 sin (B + C) sin (C + A) sin (A + B) + sin 2 A sin 2 B sin 2 C = 0$

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Solution

Given that
$cos A cos (B + C) - sin A sin (B + C) = cos A cos B cos C$
$∴ cos A [cos (B + C) - cos B cos C] = sin A sin (B + C)$
or $cos A (- sin B sin C) = sin A sin (B + C)$
$∴ sin (B + C) = \frac{- cos A sin B sin C}{sin A}$
write similar values of $sin (C + A), sin (A + B)$ and multiply and put $sin A cos A = \frac{1}{2} sin 2 A$

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