Prove that:

RHS $=4\left(\cos ^{6} x-\sin ^{6} x\right)$

$=4\left[\left(\cos ^{2} x\right)^{3}-\left(\sin ^{2} x\right)^{3}\right]$

Using the identity $a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)$, we get

$=4\left(\cos ^{2} x-\sin ^{2} x\right)\left(\cos ^{4} x+\sin ^{4} x+\sin ^{2} x \cos ^{2} x\right)$

$=4\left(\cos ^{2} x-\sin ^{2} x\right)\left(\cos ^{4} x+\sin ^{4} x+\sin ^{2} x \cos ^{2} x\right)$

$=4 \cos 2 x\left[\left(\cos ^{2} x-\sin ^{2} x\right)^{2}+2 \sin ^{2} x \cos ^{2} x+\sin ^{2} x \cos ^{2} x\right]$

$=4 \cos 2 x\left[\cos ^{2} 2 x+3 \sin ^{2} x \cos ^{2} x\right]$

$=4 \cos 2 x\left[\cos ^{2} 2 x+3\left(\frac{1-\cos ^{2} x}{2}\right)\left(\frac{1+\cos ^{2} x}{2}\right)\right]$

$=4 \cos 2 x\left[\cos ^{2} 2 x+\frac{3}{4}\left(1-\cos ^{2} 2 x\right)\right]$

$=\cos 2 x\left[4 \cos ^{2} 2 x+3\left(1-\cos ^{2} 2 x\right)\right]$

$=\cos 2 x\left[4 \cos ^{2} 2 x+3-3 \cos ^{2} 2 x\right]$

$=\cos 2 x\left[\cos ^{2} 2 x+3\right]$

$=\cos ^{3} 2 x+3 \cos 2 x=\mathrm{LHS}$

Hence proved.