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प्रश्न
Show that: \[2 \left( \sin^6 x + \cos^6 x \right) - 3 \left( \sin^4 x + \cos^4 x \right) + 1 = 0\]
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उत्तर
\[LHS = 2\left( \sin^6 x + \cos^6 x \right) - 3 \left( \sin^4 x + \cos^4 x \right) + 1\]
\[ = 2\left\{ \left( \sin^2 x \right)^3 + \left( \cos^2 \right)^3 \right\} - 3\left( \sin^4 x + \cos^4 x \right) + 1\]
\[= 2\left[ \left( \sin^2 x + \cos^2 x \right)\left( \sin^4 x - \sin^2 x \cos^2 x + \cos^4 x \right) \right] - 3\left( \sin^4 x + \cos^4 x \right) + 1\]
\[ = 2\left( \sin^4 x + \cos^4 x \right) - 2 \sin^2 x \cos^2 x - 3\left( \sin^4 x + \cos^4 x \right) + 1\]
\[ = - \left[ \sin^4 x + \cos^4 x + 2 \sin^2 x \cos^2 x \right] + 1\]
\[ = - 1 + 1 = 0 = RHS\]
\[\text{ Hence proved } .\]
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