prove that : $\cos x \cdot \cos(x-60^{\circ}) \cdot \cos(x+60^{\circ})= \frac14 \cos3x$ I should prove this trigonometric identity.
I think I should get to this point :
$\cos(3x) = 4\cos^3 x - 3\cos x $
But I don't have any idea how to do it (I tried solving $\cos(x+60^{\circ})\cos(x-60^{\circ})$ but I got nothing)

Using sum of angles identity, 
\begin{align*}
\cos x\cos (x-60^{\circ})&\cos(x+60^{\circ})\\
      & = \cos x(\cos x\cos 60^{\circ}+\sin x\sin 60^{\circ})(\cos x\cos 60^{\circ}-\sin x\sin 60^{\circ})\\ \\
      & = \cos x(\cos^2x\cos^260^{\circ}-\sin^2x\sin^260^{\circ})\\\\
      & = \cos x\left(\frac{1}{4}\cos^2x-\frac{3}{4}\sin^2x\right)\\\\
      & = \frac{1}{4}(\cos^3x-3\cos x(1-\cos^2x))\\\\
      & = \frac{1}{4}(4\cos^3x-3\cos x)
\end{align*}
