$\sin^3x + \sin^3 (x+\frac{2\pi}{3}) + \sin^3 (x+\frac{4\pi}{3})$
$=\sin^3x + \sin^3 (x-\frac{\pi}{3}) + \sin^3 (x+\frac{\pi}{3})$
$= \frac{1}{4}(3\sin x - \sin 3x) + \frac{1}{4} (3\sin(x - \frac{\pi}{3}) - \sin(3x - \pi)) +
\frac{1}{4}(\sin(x + \frac{\pi}{3}) - \sin(3x + \pi))$
$= \frac{1}{4}(3\sin x - \sin 3x + 3\sin(x - \frac{\pi}{3} ) + \sin3x - 3\sin(x + \frac{\pi}{3}) + \sin 3x)$
$= (3/4) [\sin x - \sin 3x + \sin(x - \frac{\pi}{3}) + \sin(x + \frac{\pi}{3}]$
$=\frac{3}{4}(\sin x - \sin 3x - 2 \sin x \sin\frac{\pi}{6})$
$= \frac{3}{4}(\sin x - \sin3x - \sin x)$
$= - \frac{3}{4}\sin 3x$
The following has been used in proving the above identity
$1) \sin^3 x = \frac{3\sin x - \sin3x}{4}$
$2) \sin A + \sin B = 2\sin \frac{A+B}{2} \sin \frac{A-B}{2}$
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