Exact value of polynomial at trigonometric argument 
Given that $$\cos 8\theta= 128\cos^8 \theta −256\cos^6 \theta +160 \cos^4 \theta −32\cos^2 \theta +1$$
Find the exact value of:
$$4x^4 −8x^3 +5x^2 −x$$
where $x=\cos^2 (\frac{\pi}{8})$

My try:
$$\cos 8\theta= 128\cos^8 \theta −256\cos^6 \theta +160 \cos^4 \theta −32\cos^2 \theta +1$$
$$\cos 8\theta= 32(4\cos^8 \theta -8\cos^6 \theta +5 \cos^4 \theta -\cos^2 \theta) +1$$
$$\frac{\cos 8\theta-1}{32}= 4\cos^8 \theta -8\cos^6 \theta +5 \cos^4 \theta -\cos^2 \theta$$
Let $x=\cos^2 \theta$ ? (*)
Therefore:
$$\frac{\cos 8\theta-1}{32}= 4x^4 -8x^3 +5x^2 -x$$
$\theta=\frac{\pi}{8}$
$$\frac{\cos \pi -1}{32}= 4x^4 -8x^3 +5x^2 -x$$
Which gives me : $\frac{-1}{16}$

Im not happy with my presentation, is it missing something? Specially (*)? I wrote $x=\cos^2 \theta$ when they have told $x=\cos^2 (\frac{\pi}{8})$
Can someone helps me prefect out my working. I want to improve my working. I know my result is correct. But always  in my working im not happy with my  proofs. I want to try and perfect out the method at which  I get the result.

Your approach is correct, but you should probably say that you let $\theta=\frac{\pi}{8}$ before you introduce $x$. Continuing after your third line, you could write something like this:
Let $\theta = \frac{\pi}{8}$, and denote $x=\cos^2\left(\frac{\pi}{8}\right)$. This gives
$$ \frac{\cos\pi-1}{32}=4x^4-8x^3+5x^2-x$$
so $4x^4-8x^3+5x^2-x=-\frac{1}{16}$.
