2016/077) Prove that $1+\cos 2x + \cos 4x + \cos 6x$$ = 4\cos\,x \cos 2x \cos 3x$

$1+\cos 2x + \cos 4x + \cos 6x$
= $(1+\cos 2x) + (\cos 4x + \cos 6x)$
= $(2\cos^2 x) + (\cos 4x + \cos 6x)$ using $cos 2x = 2cos^2 x - 1$
= $(2\cos^2 x) + 2\cos x  \cos 5x)$ using $\cos\, A + \cos \, B = 2(\cos \frac{A+B}{2} +  \cos \frac{B- A}{2})$
= $2\cos\,x(\cos \, x + \cos 5x)$
=$2 \cos\,x(2 \cos 2x  \cos 3x)$ using $\cos\, A + \cos\, B = 2(\cos \frac{A+B}{ 2} +  \cos \frac{B- A}{ 2})$
= $4\cos \, x\cos 2x \cos 3x$