But the root is imaginary!

$\begin{aligned} x^2+x+1 & = 0 \quad \quad \small \color{#3D99F6}{\text{Multiplying }x \text{ both sides}} \\ \implies x^3+x^2+x & = 0 \quad \quad \small \color{#3D99F6}{\text{Adding }1 \text{ both sides}} \\ x^3+\color{#3D99F6}{x^2+x + 1} & = 1 \quad \quad \small \color{#3D99F6}{\text{Note that }x^2+x + 1=0} \\ \implies x^3 & = 1 \\ x^4 & = x \\ x^5 & = x^2 \\ x^6 & = 1 \\ ... \ & = \ ... \\ \implies x^k & = x^{k \mod 3} \\ \implies x^{1999} + x^{2000} & = \color{#3D99F6}{x + x^2} \quad \quad \small \color{#3D99F6}{\text{Note that }x^2+x + 1=0} \\ & = \boxed{-1} \end{aligned}$

If $w$ is a solution of the equation $w^2+w+1=0,$ then multiplying both sides of the equation by $(w-1)$ we get the equation that $w^3-1=0,$ or, equivalently, $w^3=1$ , and therefore $w^{3k}=1.$ Then $w^{1999}+w^{2000}=w^{3(666)} w+w^{3(666)} w^2=w+w^2=-1.$

$x^2+x+1 = 0$

$x+1 = -x^2 ..... (1)$

$x=\dfrac{-1±\sqrt{3}i}{2} = e^{±\dfrac{2\pi}{3}}$

$x^{1999}+x^{2000} = x^{1999}(x+1) = x^{1999}(-x)^2$ ....... from(1)

$= -x^{2001} =- e^{±\dfrac{2\pi}{3} * 2001} = -e^{±1334\pi} = \boxed{-1}$

$x^2+x+1=0 \Rightarrow (x-1)(x^2+x+1)=0 \Rightarrow x^3=1$ This means: $x^{1999}+x^{2000}=x^{1998}(x^2+x)=(x^3)^{a}\cdot -1=1\cdot -1=-1$

Combining @Chew-Seong Cheong and @Akash Shukla 's solutions

$x^2+x+1 =0\\ x^3+x^2+x=0\\ x^3+x^2+x+1 = 1\\ \color{#D61F06}{x^3=1}$

$x^2+x+1=0 \implies \color{#3D99F6}{x+1=-x^2}$

$x^{2000}+x^{1999}\\ =x^{1999}\color{#3D99F6}{(x+1)}\\ =x^{1999}(-x^2)\\ =-x^{2001}\\ =-\color{#D61F06}{(x^3)}^{667}\\ =-1^{667}\\ =\boxed{-1}$

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Alternate method (the method I used):

$x^{2000}+x^{1999}\\ =x^{2000}+x^{1999}+x^{1998}-x^{1998}\\ =x^{1998}(x^2+x+1)-x^{1998}\\ =-x^{1998}\\ =-x^{1998}-x^{1997}-x^{1996}+x^{1997}+x^{1996}\\ =-x^{1996}(x^2+x+1)+x^{1997}+x^{1996}\\ =x^{1997}+x^{1996}\\ =x^{1997}+x^{1996}+x^{1995}-x^{1995}\\ =x^{1995}(x^2+x+1)-x^{1995}\\ =-x^{1995}\\ =-x^{1995}-x^{1994}-x^{1993}+x^{1994}+x^{1993}\\ =-x^{1993}(x^2+x+1)+x^{1994}+x^{1993}\\ =x^{1994}+x^{1993}$

Notice that $x^{2000}+x^{1999} = x^{1997}+x^{1996}=x^{1994}+x^{1993}=\ldots$

From this, we can infer that $x^n+x^{n-1} = x^{n-3}+x^{n-4}=x^{n-6}+x^{n-7}=\ldots$

We notice that the powers form an AP: $2000,1997,1994,\ldots$

Now, we look for the smallest positive term for this AP

$2000+(n-1)(-3) \geq 0\\ 2003 -3n \geq 0\\ 3n \leq 2003\\ n \leq \dfrac{2003}{3} = 667\dfrac{2}{3}$

The smallest positive term $=2000+(667-1)(-3) = 2$

From this, we know that $x^{2000}+x^{1999} = x^2+x = \boxed{-1}$

x^{2}+x+1=0 => x = ω , ω^{2} \newline x^{1999} + x^{2000} = ω + ω^{2} = -1