<s>[INST] Find $\frac{x^{3333}+x^{333}+x^{33}+x^{3}+1996}{100(x^2+x)}$ if $x^2+x+1=0$ 
Find $$\frac{x^{3333}+x^{333}+x^{33}+x^{3}+1996}{100(x^2+x)}$$ if $x^2+x+1=0$

My work so far:
1)$x^2+x=-1$, then $100(x^2+x)=-100$
2)$x^2+x+1=0$
$x=\frac{-1\pm\sqrt{3}i}{2}$
 [/INST] You can use the fact that any root of the equation $x^2 + x + 1 = 0$ is a primitive cube root of unity. Root of unity
Let's denote that root by $\omega$. Then we have $\omega^3 = 1$. So $\omega^{3333} = 1$, $\omega^{333} = 1$ and $\omega^{33} = 1$.
So your expression will turn out to be,
$$
\frac{1+1+1+1+1996}{100(-1)} = -20.
$$
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