How to prove : $\cos^32\theta + 3\cos2\theta = 4(\cos^6 \theta -\sin^6 \theta)$ How to prove : $\cos^32\theta + 3\cos2\theta = 4(\cos^6 \theta  -\sin^6 \theta)$

$4(\cos^6 \theta  -\sin^6 \theta) = 4(\cos^2 \theta  -\sin^2 \theta)(\cos^4 \theta + \cos^2 \theta\sin^2 \theta+\sin^4 \theta) = 4(\cos^2 \theta  -\sin^2 \theta)(  (\cos^2 \theta  +\sin^2 \theta)^2 - 2\cos^2 \theta\sin^2 \theta + \cos^2 \theta\sin^2 \theta) = 4(\cos^2 \theta  -\sin^2 \theta)(  1 - \cos^2 \theta\sin^2 \theta ) = 4\cos2\theta(1- \dfrac{\sin^2 2\theta}{4})  = 4\cos2\theta - \sin^2 2\theta\cos2\theta = 4\cos2\theta - (1-\cos^22\theta)\cos2\theta =  3\cos2\theta +\cos^3\theta $
