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:
$$\frac{1}{16}\left(1 - \cos(4x) + \cos(2x) - \left(\frac{\cos(6x) + \cos(-2x)}{2}\right)\right)$$
$$\frac{1}{32}\left(2 - 2\cos(4x) + 2\cos(2x) - \cos(6x) - \cos(2x)\right)$$
$$=\frac{2 + \cos(2x) - 2\cos(4x)  - \cos(6x)}{32}$$

Since $\cos(3\theta) = 4\cos^{3}(\theta) - 3\cos(\theta)$, you can also proceed backwards:
\begin{align*}
\frac{2 + \cos(2x) - 2\cos(4x) - \cos(6x)}{32} & = \frac{2 + \cos(2x) - 2(2\cos^{2}(2x) - 1) - (4\cos^{3}(2x) - 3\cos(2x))}{32}\\\\
& = \frac{-4\cos^{3}(2x) - 4\cos^{2}(2x) + 4\cos(2x) + 4}{32}\\\\
& = \frac{-\cos^{3}(2x) - \cos^{2}(2x) + \cos(2x) + 1}{8}\\\\
& = \frac{-\cos^{2}(2x)(\cos(2x) + 1) + \cos(2x) + 1}{8}\\\\
& = \frac{(1 - \cos^{2}(2x))(\cos(2x) + 1)}{8}\\\\
& = \frac{2\sin^{2}(2x)\cos^{2}(x)}{8}\\\\
& = \frac{\sin^{2}(2x)\cos^{2}(x)}{4}\\\\
& = \frac{4\sin^{2}(x)\cos^{2}(x)\cos^{2}(x)}{4}\\\\
& = \cos^{4}(x)\sin^{2}(x)
\end{align*}
Hopefully this helps!
