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)

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*$ \cos (x-60^\circ ) \times \cos ( x + 60^\circ ) = \frac{1}{2} ( \cos (2x) + \cos 120^\circ) = \frac{1}{2} \cos (2x) - \frac{1}{4}$  

*$ (\frac{1}{2} \cos (2x) - \frac{1}{4}) \times \cos x = \frac{1}{4} ( \cos (3x) + \cos (x) ) - \frac{1}{4} \cos (x) = \frac{1}{4} \cos (3x) $
