If $\alpha, \beta, \gamma, \delta$ are distinct roots of equation $x^4 + x^2 + 1 = 0$ then $\alpha^6 + \beta^6 + \gamma^6 + \delta^6$ is I tried to find discriminant first and got two roots $$\frac{-1 + \sqrt{3}i}{2}$$ and $$\frac{-1 - \sqrt{3}i}{2}$$
I tried taking $x^2 = t$ and solving equation for root $w$ and $w^2$ but got stucked, and made it more complex, any help?
Sorry if I made any silly mistake, it's been while since I practiced complex equation and finding roots. Was helping my brother with his doubts :) 

If you are going the hard way of finding explicitly the roots, then the roots and their exponential forms are
$$\alpha=-\frac{1}{2}-\frac{i\sqrt{3}}{2}=-e^{\frac{i\pi}{3}} \Rightarrow\alpha^6=e^{i2\pi}=1$$
$$\beta=\frac{1}{2}+\frac{i\sqrt{3}}{2}=e^{\frac{i\pi}{3}}\Rightarrow\beta^6=e^{i2\pi}=1$$
$$\gamma=\frac{1}{2}-\frac{i\sqrt{3}}{2}=-e^{\frac{i2\pi}{3}}\Rightarrow\gamma^6=e^{i4\pi}=1$$
$$\delta=-\frac{1}{2}+\frac{i\sqrt{3}}{2}=e^{\frac{i2\pi}{3}}\Rightarrow\delta^6=e^{i4\pi}=1$$
