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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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    $\begingroup$ Hint: multiply both sides of the given equation by $x-1$. (Since $ x \neq 1$) $\endgroup$ – nospoon Aug 28 '16 at 7:12
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    $\begingroup$ Hint: Whenever I see the constraint $x^2+x+1=0$, I immediately think of it as the third cyclotomic polynomial. Let's see. We can deduce that $$0=(x^2+x+1)(x-1)=x^3-1,$$ so $x^3=1$. I have seen you perform on the site, so I think you can take it from here :-) $\endgroup$ – Jyrki Lahtonen Aug 28 '16 at 7:12
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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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  • $\begingroup$ I was about to point out that you made a mistake by writing $x^3 = -1$. I see now that you have edited it. :-) $\endgroup$ – Prince Kumar Aug 28 '16 at 7:25
  • $\begingroup$ @PrinceKumar: Yes, I got it from raising $x=-(-1)^{1/3}$ to the power of $3$, but that yields $1$ of course. Thanks :) $\endgroup$ – barak manos Aug 28 '16 at 7:27

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