<s>[INST] 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 :) 
 [/INST] $$(-1)^3=(x^4+x^2)^3=(x^3)^4+(x^3)^2+3(x^3)^2(-1)$$
Set $x^3=y$
$$y^4-2y^2+1=0$$ whose roots are $\alpha^3$ etc.
$$(y^4-2y^2+1)^2=0$$
Set $y^2=z$
$$z^4-4z^3+\cdots+1=0$$ whose roots are $(\alpha^3)^2$ etc.
Now apply Vieta's formula
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