If $z = cis(2k\pi/5)$, $z \neq 1$, then what is $(z+1/z)^2+(z^2 + 1/z^2)^2=$? question 20, part c in the picture:
I substituted the first time as $4 \cos^2(2k \pi/5)$ and the second term as $4 \cos^2(4k \pi/5)$, and then tried writing one term in terms of the other using the identity $\cos 2a = 2 \cos^2 a- 1$. I even tried bringing in $\sin$ but I didn't get anywhere. The answer is supposed to be $3$. Can someone solve it?


Hint:
If $5t=2k\pi,5\nmid k,\cos t\ne1$
$\cos3t=\cdots==\cos2t$
The roots of 
$0=\dfrac{4\cos^3t-2\cos^2t-3\cos t+1}{\cos t-1}=4\cos^2t+2\cos t-1=0$ will be $$t=2k\pi,k\equiv\pm1,\pm2\pmod5$$
Now $z+\dfrac1z=2\cos\dfrac{2k\pi}5, \left(z+\dfrac1z\right)^2=\cdots=2\left(1+\cos\dfrac{4k\pi}5\right)$
$z^2+\dfrac1{z^2}=2\cos\dfrac{4k\pi}5,\left(z^2+\dfrac1{z^2}\right)^2=?$
