Answer: Start with left hand side,

$cos^6 \theta+sin^6 \theta=(cos^2\theta)^3+(sin^2\theta)^3=(cos^2 \theta+sin^2 \theta)^3-3cos^2 \theta sin^2 \theta (cos^2 \theta +sin^2 \theta) $

                    $=1-3cos^2 \theta sin^2 \theta$

                    $=1-3\bigg[\dfrac{1+cos 2\theta}{2}\bigg]\bigg[\dfrac{1-cos 2\theta}{2}\bigg]$

                    $=\dfrac{1}{4}\bigg [4-3(1-cos^2 2\theta)\bigg]$

                    $=\dfrac{1}{4}\bigg[4-3+3 cos^2 2 \theta \bigg]= \dfrac{1}{4}\bigg[1+3 cos^2 2 \theta\bigg]$

                    $=\dfrac{1}{4}\Bigg[1+3\Bigg({1+cos4 \theta \over 2}\Bigg)\Bigg]$

                    $= \dfrac{1}{8}\bigg[2+3+3 cos 4 \theta\bigg]=\dfrac{1}{8}\bigg[5+3 cos 4 \theta\bigg]$  (Right hand side).   Hence Proved.