This problem appeared on my latest exam and I couldnt solve it.
Knowing that w is a 36-primitive unit root solve:
$\displaystyle w^{6}+w^{12}+3 $
Sorry maybe I wasnt clear.
36-unit roots (root of unity), are all the solutions of $\displaystyle z^{36} = 1 $
Knowing that w is a 36-Primitive root of unity I have to solve
$\displaystyle w^{12} + w^6 + 3 $
In polar form we have $\displaystyle z = \cos \left( {\frac{\pi }{{18}}} \right) + i\sin \left( {\frac{\pi }{{18}}} \right).$
Now, $\displaystyle z^{12} = \cos \left( {\frac{{2\pi }}{3}} \right) + i\sin \left( {\frac{{2\pi }}{3}} \right)\quad \& \quad z^6 = \cos \left( {\frac{\pi }{3}} \right) + i\sin \left( {\frac{\pi }{3}} \right).$
Note that $\displaystyle \cos \left( {\frac{{2\pi }}{3}} \right) + \cos \left( {\frac{\pi }{3}} \right) = 0\quad \& \quad \sin \left( {\frac{{2\pi }}{3}} \right) + \sin \left( {\frac{\pi }{3}} \right) = \sqrt 3.$
Now you can finish.