Can you simply substitute?

$x^{2}+x+1=0$ $=>x^{3}=1$ $=>x^{1999}+x^{2000} = x^{1998}(x^{2}+x+1-1)=(x^{3})^{666}(-1)=-1$

From the quadratic equation, $x=\frac{-1\pm\sqrt{-3}}{2}$ , i.e. $x=e^{2\pi i/3}$ or $x=e^{4\pi i/3}$ .

Then

$x^2+x=-1$

$x^{2000}+x^{1999}=-x^{1998}=-e^{3996i/3}=-e^{2\pi i\cdot 666}=\boxed{-1}$ .

There's a much simpler solution... We have, x²+x+1=0 x+1=-x²........(1) Also, x(x+1)=-1 x+1=-1/x.....(2) So from (1) and (2) we have, -x²=-1/x x³=1

Now as asked in the question , x^1999(1+x)=x^1999×-1/x............(From eq.(2)) So, -1×x^1998 x^1998=(x^3×3×3×2×37) Now substituting value of x³=1 We get, -1×1=-1 Thus -1 is the answer...