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 :) 

$$x^4=-(x^2+1)\Rightarrow x^6=-x^4-x^2=x^2+1-x^2=1$$
So $$\sum x^6=\sum 1=1+1+1+1=4$$
Where $\displaystyle \sum x^6=\alpha^6+\beta^6+\gamma^6+\delta^6.$
