8. If real numbers $x, y$ satisfy
$$
x-3 \sqrt{x+1}=3 \sqrt{y+2}-y \text {, }
$$

then the maximum value of $x+y$ is $\qquad$.
$8.9+3 \sqrt{15}$.
Given the equation, i.e.,
$$
\begin{array}{l}
x+y=3 \sqrt{x+1}+3 \sqrt{y+2} . \\
\text { Let } u=\sqrt{x+1}, v=\sqrt{y+2} .
\end{array}
$$

Then $u \geqslant 0, v \geqslant 0, x+y=u^{2}+v^{2}-3$.
Thus, equation (1) can be transformed into
$$
\begin{array}{l}
u^{2}+v^{2}-3=3 u+3 v \\
\Leftrightarrow\left(u-\frac{3}{2}\right)^{2}+\left(v-\frac{3}{2}\right)^{2}=\frac{15}{2} .
\end{array}
$$

Therefore, point $M(u, v)$ moves on the arc $\overparen{A B}$:
$$
\left(u-\frac{3}{2}\right)^{2}+\left(v-\frac{3}{2}\right)^{2}=\left(\frac{\sqrt{30}}{2}\right)^{2}(u, v \geqslant 0)
$$

as shown in Figure 3.
Obviously, when $M$ is at point $C\left(\frac{3+\sqrt{15}}{2}, \frac{3+\sqrt{15}}{2}\right)$, $|O M|$ is maximized. At this time,
$$
\begin{array}{l}
\left(u^{2}+v^{2}\right)_{\max }=|O C|^{2}=2\left(\frac{3+\sqrt{15}}{2}\right)^{2} \\
=12+3 \sqrt{15} . \\
\text { Hence }(x+y)_{\max }=12+3 \sqrt{15}-3=9+3 \sqrt{15} .
\end{array}
$$