Prove that

$\mathrm{LHS}=4 \cos x \cos \left(\frac{\pi}{3}+x\right) \cos \left(\frac{\pi}{3}-x\right)$

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

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

$[\because 2 \cos A \cos B=\cos (A+B)+\cos (A-B)]$

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

$=2 \cos x\left[-\frac{1}{2}+\cos 2 x\right]$

$=-\cos x+2 \cos x \cos 2 x$

$=-\cos x+\cos (x+2 x)+\cos (x-2 x)$

$=-\cos x+\cos 3 x+\cos (-x)$

$=-\cos x+\cos 3 x+\cos x$

$=\cos 3 x$

RHS $=\cos 3 x$

Hence, $\mathrm{LHS}=\mathrm{RHS}$