11. (20 points) Given $x+y=1$. When do the real numbers $x$ and $y$ make $\left(x^{3}+1\right)\left(y^{3}+1\right)$ reach its maximum value?
$$
\begin{array}{l}
\text { Three, 11. Let } u=\left(x^{3}+1\right)\left(y^{3}+1\right) \\
=(x y)^{3}+(x+y)\left((x+y)^{2}-3 x y\right)+1 \\
=(x y)^{3}-3 x y+2 \triangleq t^{3}-3 t+2,
\end{array}
$$

where, $t=x y=x(1-x) \leqslant \frac{1}{4}$.
Since $u^{\prime}=3 t^{2}-3$, then
when $t0$;
when $-1<t \leqslant \frac{1}{4}$, $u^{\prime}<0$.
Therefore, $u$ reaches its maximum value at $t=-1$, i.e.,
$$
\begin{array}{l}
x y=-1, x+y=1 \\
\Rightarrow x=\frac{1 \pm \sqrt{5}}{2}, y=\frac{1 \mp \sqrt{5}}{2} .
\end{array}
$$