Key concept is to apply the angle sum identity to show the desired relationship.
$\sin(x+y) = \sin(x)\cos(y)+\cos(x)\sin(b)$
$\implies \sin(2x) = \sin(x+x) = \sin(x)\cos(x)+\cos(x)\sin(x)$
$\implies \sin(2x) = 2 \sin(x)\cos(x)$
$\cos(x + y) = \cos(x)\cos(y) - \sin(a)\sin(b)$
$\implies \cos(2x) = \cos(x+x) = \cos(x)^2 - \sin(x)^2$
$\implies \cos(3x) = \cos(1x + 2x) = \cos(2x)\cos(x) - \sin(x)\sin(2x)$
$\implies \cos(3x) = (\cos(x)^2 - \sin(x)^2)\cos(x) - \sin(x)(2 \sin(x)\cos(x))$
$\implies \cos(3x) = \cos(x)^3 - \sin(x)^2\cos(x) - 2\sin^2(x) \cos(x) $
$\implies \cos(3x) = \cos(x)^3 - 3\sin^2(x) \cos(x)$
$\implies \cos(x)^3 = \cos(3x) + 3\sin^2(x) \cos(x)$
$\sin^2(x) + \cos^2(x) = 1$
$\implies \sin^2 = 1 - \cos^2(x)$
$\implies \cos(x)^3 = \cos(3x) + 3 (1 - \cos^2(x)) \cos(x)$
$\implies \cos(x)^3 = \cos(3x) + 3\cos(x) - 3\cos^3(x) $
$\implies 4 \cos(x)^3 = \cos(3x) + 3\cos(x)$
$\implies \cos(x)^3 = \frac{1}{4} (\cos(3x) + 3\cos(x))$