Answer
See the steps.
Work Step by Step
$LHS =\cos{3\theta} = \cos{(2\theta +\theta)} = \cos{2\theta} \cos{\theta} - \sin{2\theta} \sin{\theta}$
$\cos{3\theta} = (2\cos^2 {\theta}-1)(\cos{\theta}) - 2 \sin^2{\theta} \cos{\theta}$
$\cos{3\theta} = 2\cos^3{\theta} - \cos{\theta} - 2 (1-\cos^2{\theta}) \cos{\theta}$
$ = 2 \cos^3{\theta} -\cos{\theta} -2\cos{\theta} + 2 \cos^3{\theta} $
$LHS = 4 \cos^3{\theta} -3 \cos{\theta} = RHS$
