Prove the identity: $2 \left(\sin^6 \alpha + \cos^6 \alpha\right) - 3 \left(\sin^4 \alpha + \cos^4 \alpha\right) + 1 = 0$
$\begin{aligned}
LHS & = 2 \left(\sin^6 \alpha + \cos^6 \alpha\right) - 3 \left(\sin^4 \alpha + \cos^4 \alpha\right) + 1 \\\\
& = 2 \left[\left(\sin^2 \alpha\right)^3 + \left(\cos^2 \alpha\right)^3\right] - 3 \sin^4 \alpha - 3 \cos^4 \alpha + 1 \\\\
& = 2 \left[\left(\sin^2 \alpha + \cos^2 \alpha\right) \left(\sin^4 \alpha - \sin^2 \alpha \cos^2 \alpha + \cos^4 \alpha\right)\right] \\
& \hspace{6cm} - 3 \sin^4 \alpha - 3 \cos^4 \alpha + 1 \\\\
& = 2 \sin^4 \alpha - 2 \sin^2 \alpha \cos^2 \alpha + 2 \cos^4 \alpha - 3 \sin^4 \alpha - 3 \cos^4 \alpha + 1 \\\\
& = - \sin^4 \alpha - 2 \sin^2 \alpha \cos^2 \alpha - \cos^4 \alpha + 1 \\\\
& = - \left[\sin^4 \alpha + 2 \sin^2 \alpha \cos^2 \alpha + \cos^4 \alpha\right] + 1 \\\\
& = - \left[\sin^2 \alpha + \cos^2 \alpha\right]^2 + 1 \\\\
& = - 1 + 1 \\\\
& = 0 = RHS
\end{aligned}$
Hence proved.