7. Let non-zero distinct complex numbers $x, y$ satisfy $x^{2}+x y+$ $y^{2}=0$. Then the value of the expression
$$
\left[\frac{x y}{(x+y)(x-y)^{2}}\right]^{2000}\left(x^{2006}+y^{2006}\right)
$$

is $\qquad$ .
$=7 .-\frac{1}{3^{2000}}$.
From $x 、 y$ being non-zero and distinct, we get $\left(\frac{x}{y}\right)^{2}+\frac{x}{y}+1=0$, then $\frac{x}{y}=\omega$ is a complex cube root of unity. Therefore,
$$
\begin{array}{l}
\text { Original expression }=\left[\frac{\omega}{(1+\omega)(1-\omega)^{2}}\right]^{2006}\left(\omega^{2006}+1\right) \\
=\left[\frac{\omega}{\left(-\omega^{2}\right)(-3 \omega)}\right]^{2006}\left(\omega^{2}+1\right) \\
=\frac{\omega^{2}}{3^{2006}}(-\omega)=-\frac{1}{3^{2006}} .
\end{array}
$$