^3x+ ^3 (2 3 +x )+ ^3 (4 3 +x )=-3 4 3x

Question

$\sin^3\text{x}+\sin^3\big(\frac{2\pi}{3}+\text{x}\big)+\sin^3\big(\frac{4\pi}{3}+\text{x}\big)=-\frac{3}{4}\sin3\text{x}$

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Answer

$\text{LHS}=\sin^3\text{x}+\sin^3\Big(\frac{2\pi}{3}+\text{x}\Big)+\sin^3\Big(\frac{4\pi}{3}+\text{x}\Big)$ $\Big\{$ We know that $\sin^3\text{x}=\frac{3\sin\text{x}-\sin3\text{x}}{4}\Big\}$ $=\Big(\frac{3\sin\text{x}-\sin3\text{x}}{4}\Big)+\Bigg\{\frac{3\sin\big(\frac{2\pi}{4}+\text{x}\big)-\sin3\big(\frac{2\pi}{3}+\text{x}\big)}{4}\Bigg\}\\+\Bigg\{\frac{3\sin\big(\frac{4\pi}{3}+\text{x}\big)-\sin3\big(\frac{4\pi}{3}+\text{x}\big)}{4}\Bigg\}$ $=\Big[\frac{3\sin\text{x}-\sin3\text{x}}{4}\Big]+\Bigg[\frac{3\sin\big[\pi\big(\frac{2\pi}{3}+\text{x}\big)\big]-\sin(2\pi+3\text{x})}{4}\Bigg]\\+\Bigg[\frac{3\sin\big[\pi\big(\frac{\pi}{3}+\text{x}\big)\big]-\sin(4\pi+3\text{x})}{4}\Bigg]$ $=\frac{1}{4}\Big\{[3\sin\text{x}-\sin3\text{x}]+\Big[3\sin\Big(\frac{\pi}{3}-\text{x}\Big)-\sin3\text{x}\Big]\\-\Big[3\sin\Big(\frac{\pi}{3}+\text{x}\Big)+\sin3\text{x}\Big]\Big\}$ $=\frac{1}{4}\Big[3\sin\text{x}-\sin3\text{x}+3\sin\Big(\frac{\pi}{3}-\text{x}\Big)-3\sin\Big(\frac{\pi}{3}+\text{x}\Big)-\sin3\text{x}\sin3\text{x}\Big]$ $=\frac{1}{4}\Big[3\sin\text{x}-3\sin3\text{x}+3\Big(\sin\Big(\frac{\pi}{3}-\text{x}\Big)-\sin\Big(\frac{\pi}{3}+\text{x}\Big)\Big)\Big]$ $=\frac{1}{4}\Bigg[3\sin\text{x}-3\sin3\text{x}+3\Bigg\{2\cos\frac{\frac{\pi}{3}-\text{x}+\frac{\pi}{3}+\text{x}}{2}\sin\frac{\frac{\pi}{3}-\text{x}\frac{\pi}{3}-\text{x}}{2}\Bigg\}\Bigg]$ $=\frac{1}{4}\Big[3\sin\text{x}-3\sin2\text{x}+6\cos\frac{\pi}{3}\sin(-\text{x})\Big]$ $\frac{1}{4}[3\sin\text{x}-3\sin3\text{x}-3\sin\text{x}]$ $=-\frac{3}{4}\sin3\text{x}$ $=\text{RHS}$ $\text{LHS}=\text{RHS}$