Question

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

Answer 1

$$\begin{gathered} \mathrm{LHS}={\sin }^{3}x+{\sin }^3\left(\frac{2\mathrm{\pi }}{3}+x\right)+{\sin }^3\left(\frac{4\mathrm{\pi }}{3}+x\right) \\ =\frac{3\sin x-\sin 3x}{4}+\frac{3\sin \left({\displaystyle \frac{2\mathrm{\pi }}{3}}+x\right)-\sin 3\left(\frac{2\mathrm{\pi }}{3}+x\right)}{4}+\frac{3\sin \left({\displaystyle \frac{4\mathrm{\pi }}{3}}+x\right)-\sin 3\left(\frac{4\mathrm{\pi }}{3}+x\right)}{4} \\ \left[{\sin }^3\mathrm{\theta }=\frac{3\mathrm{sin\theta }-\sin 3\mathrm{\theta }}{4}\right] \\ =\frac{3\sin x-\sin 3x}{4}+\frac{3\sin \left\{\mathrm{\pi }-\left(\frac{2\mathrm{\pi }}{3}+x\right)\right\}-\sin \left(2\mathrm{\pi }+3x\right)}{4}+\frac{3\sin \left\{\mathrm{\pi }+\left(\frac{\mathrm{\pi }}{3}+x\right)\right\}-\sin \left(4\mathrm{\pi }+3x\right)}{4} \\ =\frac{1}{4}\left[\left(3\sin x-\sin 3x\right)+\left\{3\sin \left(\frac{\mathrm{\pi }}{3}-x\right)-\sin 3\mathrm{x}\right\}-\left\{3\sin \left(\frac{\mathrm{\pi }}{3}+x\right)+\sin 3x\right\}\right] \\ =\frac{1}{4}\left[3\sin x-\sin 3x+3\sin \left(\frac{\mathrm{\pi }}{3}-x\right)-3\sin \left(\frac{\mathrm{\pi }}{3}+x\right)-\sin 3x-\sin 3x\right] \end{gathered}$$

$$\begin{gathered} =\frac{1}{4}\left[3\sin x-3\sin 3x+3\left\{\sin \left(\frac{\mathrm{\pi }}{3}-x\right)-\sin \left(\frac{\mathrm{\pi }}{3}+x\right)\right\}\right] \\ =\frac{1}{4}\left[3\sin x-3\sin 3x+3\left\{2\cos \frac{{\displaystyle \frac{\mathrm{\pi }}{3}}-x+{\displaystyle \frac{\mathrm{\pi }}{3}}+x}{2}\sin \frac{{\displaystyle \frac{\mathrm{\pi }}{3}}-x-{\displaystyle \frac{\mathrm{\pi }}{3}}-x}{2}\right\}\right] \\ \left[\because \mathrm{sinC}-\mathrm{sinD}=2\cos \frac{\mathrm{C}+\mathrm{D}}{2}\sin \frac{\mathrm{C}-\mathrm{D}}{2}\right] \\ =\frac{1}{4}\left[3\sin x-3\sin 3x+6\cos \frac{\mathrm{\pi }}{3}\sin \left(-x\right)\right] \\ =\frac{1}{4}\left[3\sin x-3\sin 3x-3\sin x\right] \\ =-\frac{3}{4}\sin x \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$