8. (10th "Hope Cup" Invitational Competition Question) Let $x_{1}, x_{2}$ be the roots of the equation $x^{2}+x+1=0$, then the value of the sum $\frac{x_{1}}{x_{2}}+\left(\frac{x_{1}}{x_{2}}\right)^{2}+\left(\frac{x_{1}}{x_{2}}\right)^{3}+\cdots+\left(\frac{x_{1}}{x_{2}}\right)^{1998}$ is $\qquad$ $ـ$.
8. Fill 0. Reason: Since $1998=3 \cdot 666$, the sum can be transformed into the sum of 666 three-term expressions. These three-term expressions are
$$
\begin{array}{l}
\frac{x_{1}}{x_{2}}+\left(\frac{x_{1}}{x_{2}}\right)^{2}+\left(\frac{x_{1}}{x_{2}}\right)^{3},\left(\frac{x_{1}}{x_{2}}\right)^{4}+\left(\frac{x_{1}}{x_{2}}\right)^{5}+\left(\frac{x_{1}}{x_{2}}\right)^{6}, \\
\left(\frac{x_{1}}{x_{2}}\right)^{1993}+\left(\frac{x_{1}}{x_{2}}\right)^{1994}+\left(\frac{x_{1}}{x_{2}}\right)^{1995},\left(\frac{x_{1}}{x_{2}}\right)^{1996}+\left(\frac{x_{1}}{x_{2}}\right)^{1997}+\left(\frac{x_{1}}{x_{2}}\right)^{1998} .
\end{array}
$$

For each of these three-term expressions, factor out the lowest power of $\frac{x_{1}}{x_{2}}$, then the sum becomes $\left[\frac{x_{1}}{x_{2}}+\left(\frac{x_{1}}{x_{2}}\right)^{4}+\cdots+\left(\frac{x_{1}}{x_{2}}\right)^{1996}\right] \cdot$ $\left[1+\frac{x_{1}}{x_{2}}+\left(\frac{x_{1}}{x_{2}}\right)^{2}\right]$

Since $x_{1}$ and $x_{2}$ are the roots of the equation $x^{2}+x+1=0$, then $x_{1}+x_{2}=-1, x_{1} \cdot x_{2}=1$, which means $\left(x_{1}+x_{2}\right)^{2}-x_{1} x_{2}=0$, thus $1+\frac{x_{1}}{x_{2}}+\left(\frac{x_{1}}{x_{2}}\right)^{2}=\frac{x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}}{x_{2}^{2}}=\frac{\left(x_{1}+x_{2}\right)^{2}-x_{1} x_{2}}{x_{2}^{2}}=0$, hence the value of the sum is 0.

Or by solving the equation $x^{2}+x+1=0$, we get $x_{1}=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}, x_{2}=-\frac{1}{2}-\frac{\sqrt{3}}{2} \mathrm{i}$, thus $\frac{x_{1}}{x_{2}}=-\frac{1}{2}-\frac{\sqrt{3}}{2} \mathrm{i},\left(\frac{x_{1}}{x_{2}}\right)^{2}=$ $-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i},\left(\frac{x_{1}}{x_{2}}\right)^{3}=1$. Therefore, $\frac{x_{1}}{x_{2}}+\left(\frac{x_{1}}{x_{2}}\right)^{2}+\left(\frac{x_{1}}{x_{2}}\right)^{3}=0$, so the original expression's value is 0.