Compute $\cos(20^\circ)\cos(40^\circ)\cos(80^\circ)$ using roots of unity . 
Compute $\cos(20^\circ)\cos(40^\circ)\cos(80^\circ)$.

The hint says to take $\zeta=\text {cos}(20^\circ)+ \text {i sin} (20^\circ)$ and then asks us to write $\cos(20^\circ)$  is terms of $\zeta$ .
Now, I took $\zeta=\text {cos}(20^\circ)+ \text {i sin} (20^\circ)$ be the 18 root of unit. So $\zeta^2=\text {cos}(40^\circ)+ \text {i sin} (40^\circ)$ and $\zeta^4=\text {cos}(80^\circ)+ \text {i sin} (80^\circ)$.
But I didn't understand , how to write $\cos(20^\circ)$  is terms of $\zeta$ .
Any hints ? Thanks in advance !

Try to substitute the following:
$$
\begin{align}
2\cos{(20)}&=\zeta-\zeta^{8}\\
2\cos{(40)}&=\zeta^{2}-\zeta^{7}\\
2\cos{(80)}&=\zeta^{4}-\zeta^{5}\\
\\
\zeta^{9}&=-1
\end{align}
$$
Then use the equation
$$
\begin{align}
\zeta^{8}-\zeta^{7}+\zeta^{6}-\zeta^{5}+\zeta^{4}-\zeta^{3}+\zeta^{2}-\zeta+1&=0
\end{align}
$$
