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Prove that: $2(\sin^6 \theta + \cos^6 \theta) -3(\sin^4 \theta + \cos^4 \theta)+1=0$.
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LHS = $$\begin{aligned}& 2(\sin^6\theta + \cos^6\theta) - 3(\sin^4\theta + \cos^4\theta) + 1 \\ & = 2[(\sin^2\theta)^3 + (\cos^2\theta)^3] - 3(\sin^4\theta + \cos^4\theta) + 1 \\ & = 2[(\sin^2\theta + \cos^2\theta)(\sin^4\theta - \sin^2\theta \cos^2\theta + \cos^4\theta)] - 3(\sin^4\theta + \cos^4\theta) + 1 \\ & = 2[\sin^4\theta + \cos^4\theta - \sin^2\theta \cos^2\theta] - 3(\sin^4\theta + \cos^4\theta) + 1 \\ & = -[\sin^4\theta + \cos^4\theta + 2 \sin^2\theta \cos^2\theta] + 1 \\ & = -(\sin^2\theta + \cos^2\theta)^2 + 1 \\ & = -1 + 1 = 0\end{aligned}$$