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 $\qquad$
$$
\text { 8. } 9+3 \sqrt{15} \text {. }
$$

Given the equation:
$$
\begin{array}{l}
x+y=3 \sqrt{x+1}+3 \sqrt{y+2} . \\
\text { Let } u=\sqrt{x+1}, v=\sqrt{y+2} .
\end{array}
$$

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

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