Mighty Big Powers!

See: De Moivre's Theorem

$(\cos(\pi/3)+i\sin(\pi/3))^{2016}+(\cos(-\pi/3)+i\sin(-\pi/3))^{2016}$ $=(\cos(2016\pi/3)+i\sin(2016\pi/3))+(\cos(2016\pi/3)+i\sin(-2016\pi/3))$ $\color{#D61F06}{\text{(Using De Moivre's Rule)}}$ $=2\cos(672\pi)=\boxed 2$

Since $\omega=-\frac{1}{2}+\frac{\sqrt{-3}}{2}$ and $\omega^2=-\frac{1}{2}-\frac{\sqrt{-3}}{2}$ are the non real cube roots of unity and $-\omega=\frac{1}{2}-\frac{\sqrt{-3}}{2}$ and $-\omega^2=\frac{1}{2}+\frac{\sqrt{-3}}{2}$ and $\omega^3=1$ . Therefore:

$\left( \frac{1}{2}+\frac{\sqrt{-3}}{2} \right)^{2016}+\left(\frac{1}{2}-\frac{\sqrt{-3}}{2} \right)^{2016} =$ $\left( -\omega^2 \right)^{2016}+\left(-\omega \right)^{2016} =$

$\left( \omega^2 \right)^{2016}+\left(\omega \right)^{2016} =$ $\left( \omega^2 \right)^{3 \times 672}+\left(\omega \right)^{3 \times 672} =$ $\left( \omega^3 \right)^{2 \times 672}+\left(\omega^3 \right)^{672} =$ $\left( 1\right)^{2 \times 672}+\left(1 \right)^{672} =1+1=\boxed2$

Just use De moivre's theorem.