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 :) 

Hint: Recall that $(a-b)(a^2+ab+b^2) = a^3-b^3$
Multiplying $$x^4 + x^2 + 1 = 0$$ with $x^2-1$ we get $$x^6-1=0$$ so $x^6=1$ and now is easy to finish...
