If $x^2+x+1 = 0$ then find the value of $x^{1999}+x^{2000}$.
I first tried finding the solution of the given equation and then substituting it in the expression whose value we have to find but I wasn't able to simplify it.
In a different approach I moved the terms around a bit and arrived at $x^3 = 1$. But wouldn't that mean that $x = 1$ (which is clearly not possible since it wouldn't satisfy the given equation)? Any help would be appreciated.
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$\begingroup$Hint: As you have correctly observed, we can deduce that $x^3 = 1$. Now, note that $$ x^{1999} + x^{2000} = (x^{3})^{666}(x + x^2) $$
$\endgroup$ 5 $\begingroup$$x^{1998}(x^2+x)=-x^{1998}=-(x^{3})^{666}=-1$
$\endgroup$ $\begingroup$There is nice relation here $$x^{2000}+x^{1999}=-x^{1998}=x^{1999}+x^{1997}$$ which is why $x^3=1$ So $$x^{1998}(x+x^2)=({x^{3}})^{666}(-1)=-1$$
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