3. Given two non-zero complex numbers $x$ and $y$ whose sum of cubes is zero. Then $\left(\frac{x}{x-y}\right)^{2000}+\left(\frac{y}{x-y}\right)^{2000}$ is $\qquad$
3. $2^{-1999}$ or -1.

Let $x=w y$. Then $w^{3}+1=0$.
Solving, we get $w=-1$ or $w^{2}-w+1=0$.
When $w=-1$,
$$
\begin{array}{l}
\text { the expression }=\left(\frac{w}{w-1}\right)^{2000}+\left(\frac{1}{w-1}\right)^{2000} \\
=2^{-2000}+2^{-2000}=2^{-1999} .
\end{array}
$$

When $w^{2}-w+1=0$,
$$
\begin{array}{l}
\text { the expression }=\left(\frac{w^{2}}{w^{2}-w}\right)^{2000}+\left(\frac{w}{w^{2}-w}\right)^{2000} \\
=\left(\frac{w^{2}}{-1}\right)^{2000}+\left(\frac{w}{-1}\right)^{2000}=w^{4000}+w^{2000} \\
=\left(w^{3}\right)^{1333} w+\left(w^{3}\right)^{666} w^{2} \\
=(-1)^{1333} w+(-1)^{666} w^{2}=-w+w^{2}=-1 .
\end{array}
$$