<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] $$
(x^2-1)(x^4+x^2+1) = x^6 - 1
$$
$x^6-1=0$ if and only if $x$ is one of the $6$th roots of $1.$
$(x^2-1)(x^4+x^2+1)=0$ if and only if either $x^2-1=0,$ in which case $x=\pm 1,$ or $x^4+x^2+1=0,$ which fails to hold if $x=\pm 1,$ since $(\pm 1)^4+(\pm 1)^2 + 1 = 3\ne0.$ Thus the roots of $x^4+x^2+1=0$ are the $6$th roots of $1$ other than $\pm 1.$ Their $6$th powers are equal to $1,$ so the sum of their $6$th powers is $4.$
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