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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 = $2[(\sin^2 \theta + \cos^2 \theta)^3 - 3 \sin^2 \theta \cos^2 \theta(\sin^2 \theta + \cos^2 \theta)] - 3[(\sin^2 \theta + \cos^2 \theta)^2 - 2 \sin^2 \theta \cos^2 \theta] + 1$
$= 2[1 - 3 \sin^2 \theta \cos^2 \theta] - 3[1 - 2 \sin^2 \theta \cos^2 \theta] + 1$
$= 2 - 6 \sin^2 \theta \cos^2 \theta - 3 + 6 \sin^2 \theta \cos^2 \theta + 1 = 0 = \text{RHS}$
$= 2[1 - 3 \sin^2 \theta \cos^2 \theta] - 3[1 - 2 \sin^2 \theta \cos^2 \theta] + 1$
$= 2 - 6 \sin^2 \theta \cos^2 \theta - 3 + 6 \sin^2 \theta \cos^2 \theta + 1 = 0 = \text{RHS}$