sin

$\mathrm{LHS}=\sin ^{3} x+\sin ^{3}\left(\frac{2 \pi}{3}+x\right)+\sin ^{3}\left(\frac{4 \pi}{3}+x\right)$

$=\frac{3 \sin x-\sin 3 x}{4}+\frac{3 \sin \left(\frac{2 \pi}{3}+x\right)-\sin 3\left(\frac{2 \pi}{3}+x\right)}{4}+\frac{3 \sin \left(\frac{4 \pi}{3}+x\right)-\sin 3\left(\frac{4 \pi}{3}+x\right)}{4}$

$\left[\sin ^{3} \theta=\frac{3 \sin \theta-\sin 3 \theta}{4}\right]$

$=\frac{3 \sin x-\sin 3 x}{4}+\frac{3 \sin \left\{\pi-\left(\frac{2 \pi}{3}+x\right)\right\}-\sin (2 \pi+3 x)}{4}+\frac{3 \sin \left\{\pi+\left(\frac{\pi}{3}+x\right)\right\}-\sin (4 \pi+3 x)}{4}$

$=\frac{1}{4}\left[(3 \sin x-\sin 3 x)+\left\{3 \sin \left(\frac{\pi}{3}-x\right)-\sin 3 x\right\}-\left\{3 \sin \left(\frac{\pi}{3}+x\right)+\sin 3 x\right\}\right]$

$=\frac{1}{4}\left[3 \sin x-\sin 3 x+3 \sin \left(\frac{\pi}{3}-x\right)-3 \sin \left(\frac{\pi}{3}+x\right)-\sin 3 x-\sin 3 x\right]$

$=\frac{1}{4}\left[3 \sin x-3 \sin 3 x+3\left\{\sin \left(\frac{\pi}{3}-x\right)-\sin \left(\frac{\pi}{3}+x\right)\right\}\right]$

$=\frac{1}{4}\left[3 \sin x-3 \sin 3 x+3\left\{2 \cos \frac{\frac{\pi}{3}-x+\frac{\pi}{3}+x}{2} \sin \frac{\frac{\pi}{3}-x-\frac{\pi}{3}-x}{2}\right\}\right]$

$\left[\because \sin \mathrm{C}-\sin \mathrm{D}=2 \cos \frac{\mathrm{C}+\mathrm{D}}{2} \sin \frac{\mathrm{C}-\mathrm{D}}{2}\right]$

$=\frac{1}{4}\left[3 \sin x-3 \sin 3 x+6 \cos \frac{\pi}{3} \sin (-x)\right]$

$=\frac{1}{4}[3 \sin x-3 \sin 3 x-3 \sin x]$

$=-\frac{3}{4} \sin x$

= RHS

Hence proved.