Find $\frac{x^{3333}+x^{333}+x^{33}+x^{3}+1996}{100(x^2+x)}$ if $x^2+x+1=0$

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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)$x^2+x+1=0$

$x=\frac{-1\pm\sqrt{3}i}{2}$

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3 Answers

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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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$$x^2=-x-1 \Rightarrow x^3=-x^2-x=(x+1)-x=1 \ \ \& \ \ \Rightarrow x^{33}=(x^3)^{11}=1$$ $$\& \ \ \ x^{333}=(x^3)^{111}=x^{3333}=(x^3)^{1111}=x^3=1.$$ Also: $$x^{2}+x=x(x+1)=x(-x^{2})=-x^3=-1.$$ So: $$\dfrac{x^{3333}+x^{333}+x^{33}+x^{3}+1996}{100(x^2+x)}=\dfrac{1+1+1+1+1996}{100\times (-1)}=\dfrac{2000}{-100}=-20.$$

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$x^2+x+1=0\implies$

$x=-(-1)^{1/3},(-1)^{2/3}\implies$

$x^3=1\implies$

$\cfrac{x^{3333}+x^{333}+x^{33}+x^{3}+1996}{-100}=\cfrac{1^{1111}+1^{111}+1^{11}+1^{1}+1996}{-100}=-20$

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