A right substitution

$\large Let {\huge f}(n)= x^n + \dfrac 1 x^n.~~~If~x^2 + x+ 1 = 0,~~we~have~\\\large A={\huge f}\left (~~ n\equiv 1 \pmod3 ~~~\right )=-1 ,\\\large B= {\huge f}\left (~~n\equiv 2 \pmod3~~\right )=-1\\\large C={\huge f}\left (~~n\equiv 0 \pmod3~~\right )=2.\\\therefore A^2+B^2+C^2=1+1+4=6 \\~~\\ 1000=333*3+1.\\ \left \{~~ \left ( x + \frac 1 x \right )^2 + \left ( x^2 + \frac 1 {x^2} \right )^2 + \left ( x^3 + \frac 1 {x^3} \right )^2~~\right \} +\\ \left \{ \left ( x^4 + \frac 1 {x^4} +\right )^2+ \left ( x^5 + \frac 1 {x^5} \right )^2 + \left ( x^6 + \frac 1 {x^6} +\right )^2 \right \} \ldots + \left ( x^{1000} + \frac {1} {x^{1000}} \right )^2\\=(A^2+B^2+C^2)+(A^2+B^2+C^2)+.....+A^2\\=333*6+1=1999.$

Showing the solution suggested by Miraj Shah .

If $x^2+x+1=0\quad \Rightarrow x^3+x^2+x =0 \quad \Rightarrow x^3+x^2+x + 1 = 1$ $\Rightarrow x^3 = 1$ . Therefore, $x$ is the cubic root of $1$ .

$\Rightarrow x^3 = 1 = e^{2\pi i} \quad \Rightarrow x = e^{\frac{2\pi}{3} i} = \cos{\frac{2\pi}{3}} + i\sin{\frac{2\pi}{3}}$

$\begin{aligned} \Rightarrow \displaystyle \sum_{k=1}^{1000} {\left(x^k+\dfrac{1}{x^k}\right)^2} & = \sum_{k=1}^{1000} {\left(e^{\frac{2k\pi}{3}i} +e^{-\frac{2k\pi}{3}i} \right)^2} = \sum_{k=1}^{1000} {\left(2\cos{\frac{2k\pi}{3}} \right)^2} \\ & = \sum_{k=1}^{1000} {4\cos^2{\frac{2k\pi}{3}}} \\ & = 4 \left( \frac{1}{4} + \frac{1}{4} + 1 + \frac{1}{4} + \frac{1}{4} + 1 + ...+ \frac{1}{4} + 1 + \frac{1}{4} \right) \\ & = 4 \left( 333\left[ \frac{1}{4} + \frac{1}{4} + 1\right] + \frac{1}{4} \right)\\ & = 333(6)+1 = \boxed{1999} \end{aligned}$

$x^{2}+x+1=0$ $\Rightarrow$ $x=\omega$ . $\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\$ $(x+\frac{1}{x})^{2}+(x^{2}+\frac{1}{x^{2}})^{2}+. . . . . . . . . .+(x^{1000}+\frac{1}{x^{1000}})^{2}=(1+1+4)+(1+1+4)+ . . . . . . . .$ (333 times) $+1=6\times333+1=\boxed{1999}$ .

Actually it is just the point of x^4=x^2+2x+1=x,
that means x^3=1, and expand the expression and one part will become (x^2+1+x)+ (x^2+1+x)+( x^2+1+x)+......+(x^2+1+x)+x^2=x^2; another part is 1/(x^2); the rest are just 2000; so totally it is x^2+1/x^2+2000=1999

lol, you guys think of it with a complex way but sometimes simpleness is beauty but sometimes complexity is also beauty that is math

The above equation's roots are nothing but complex cube root of 1. This info can be used to solve the question even faster