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