Rewriting $\cos^4 x \sin^2 x $ with exponent no higher than $1$ I'm having some trouble finishing this one off.

Rewrite with exponent no higher than $1$:
$$\cos^4 x \sin^2 x$$

The answer is:

$$\frac{2 + \cos(2x) - 2\cos(4x) - \cos(6x)}{32}$$

So I started like this:
$$\cos^4 x \sin^2 x = \frac{1+\cos(2x)}{2}\frac{1+\cos(2x)}{2}\frac{1-\cos(2x)}{2}$$
$$= \frac{1}{8}\left(\{1+\cos(2x)\}\{1^2 - \cos^2(2x)\}\right)$$
$$\frac{1}{8}\left(\{1 + \cos(2x)\}\sin^2(2x)\right)$$
$$\frac{1}{16}\left(1 + \cos(2x)\{1-\cos(4x)\}\right)$$
Now this is where I start to get lost:
$$\frac{1}{16}\left(1 - \cos(4x) + \cos(2x) - \cos(2x)\cos(4x) \right)$$
I really can't find a way from here - I try this, but not sure if this is the right path.
$$\require{cancel} \cancel{\frac{1}{16}\left(1 - \cos(4x) + \cos(2x)\{1 - \cos(4x)\} \right)}$$
Completing thanks to help below:
$$\frac{1}{16}\left(1 - \cos(4x) + \cos(2x) - \left(\frac{\cos(6x) + \cos(-2x)}{2}\right)\right)$$
$$\frac{1}{32}\left(2 - 2\cos(4x) + 2\cos(2x) - \cos(6x) - \cos(2x)\right)$$
$$=\frac{2 + \cos(2x) - 2\cos(4x)  - \cos(6x)}{32}$$

$\cos x = \frac {e^{ix} + e^{-ix}}{2}\\
\sin x = \frac {e^{ix} - e^{-ix}}{2i}\\
\cos^4 x\sin^2 x = \frac {(e^{ix} + e^{-ix})^4(e^{ix} - e^{-ix})^2}{-64}\\
\frac {(e^{4ix} + 4e^{2ix} + 6 + 4e^{-2ix} + e^{-4ix})(e^{2ix} - 2 + e^{-2ix})}{-64}\\
\frac {e^{6ix} + 2e^{4ix} - e^{2ix} -4 - e^{-2ix}+2e^{-4ix} + e^{-6ix}}{-64}\\
\frac {\cos 6x + 2\cos 4x - \cos 2x - 2}{-32}\\
\frac {-\cos 6x - 2\cos 4x + \cos 2x+2}{32}$
