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प्रश्न
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उत्तर
\[LHS = \sin^3 x + \sin^3 \left( \frac{2\pi}{3} + x \right) + \sin^3 \left( \frac{4\pi}{3} + x \right)\]
\[ = \frac{3\text{ sin } x - \sin3x}{4} + \frac{3\sin\left( \frac{2\pi}{3} + x \right) - \sin3\left( \frac{2\pi}{3} + x \right)}{4} + \frac{3\sin\left( \frac{4\pi}{3} + x \right) - \sin3\left( \frac{4\pi}{3} + x \right)}{4} \]
\[ \left[ \sin^3 \theta = \frac{3sin\theta - \sin3\theta}{4} \right]\]
\[ = \frac{3\text{ sin } x - \sin3x}{4} + \frac{3\sin\left\{ \pi - \left( \frac{2\pi}{3} + x \right) \right\} - \sin\left( 2\pi + 3x \right)}{4} + \frac{3\sin\left\{ \pi + \left( \frac{\pi}{3} + x \right) \right\} - \sin\left( 4\pi + 3x \right)}{4}\]
\[ = \frac{1}{4}\left[ \left( 3\text{ sin } x - \sin3x \right) + \left\{ 3\sin\left( \frac{\pi}{3} - x \right) - \sin3x \right\} - \left\{ 3\sin\left( \frac{\pi}{3} + x \right) + \sin3x \right\} \right]\]
\[ = \frac{1}{4}\left[ 3\text{ sin } x - \sin3x + 3\sin\left( \frac{\pi}{3} - x \right) - 3\sin\left( \frac{\pi}{3} + x \right) - \sin3x - \sin3x \right]\]
\[= \frac{1}{4}\left[ 3\text{ sin } x - 3\sin3x + 3\left\{ \sin\left( \frac{\pi}{3} - x \right) - \sin\left( \frac{\pi}{3} + x \right) \right\} \right]\]
\[ = \frac{1}{4}\left[ 3\text{ sin } x - 3\sin3x + 3\left\{ 2\cos\frac{\frac{\pi}{3} - x + \frac{\pi}{3} + x}{2}\sin\frac{\frac{\pi}{3} - x - \frac{\pi}{3} - x}{2} \right\} \right]\]
\[ \left[ \because sinC - sinD = 2\cos\frac{C + D}{2}\sin\frac{C - D}{2} \right]\]
\[ = \frac{1}{4}\left[ 3\text{ sin } x - 3\sin3x + 6\cos\frac{\pi}{3}\sin\left( - x \right) \right]\]
\[ = \frac{1}{4}\left[ 3\text{ sin } x - 3\text{ sin } 3x - 3\text{ sin } x \right]\]
\[ = - \frac{3}{4}\text{ sin } x\]
\[ = RHS\]
\[\text{ Hence proved } .\]
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