9. (2007 Croatian Mathematical Competition) Find all integer solutions to the equation $x^{3}+11^{3}=y^{3}$.
9. From observation, it is easy to see that $y > x$, i.e., $y - x > 0$.

The original equation is equivalent to
$$
11^{3} = y^{3} - x^{3} = (y - x)\left(y^{2} + x y + x^{2}\right).
$$

Since 11 is a prime number, $y - x$ can only be $1, 11, 11^{2}$, or $11^{3}$.
If $y - x = 1$, then
$$
x^{2} + x y + y^{2} = 11^{3} = 3 x^{2} + 3 x + 1.
$$

This equation has no integer solutions.
If $y - x = 11$, then
$$
11^{2} = x^{2} + x y + y^{2} = 3 x^{2} + 3 \times 11 x + 11^{2}.
$$

Solving this, we get $x = 0, y = 11$ or $x = -11, y = 0$.
If $y - x = 11^{2}$, then
$$
11 = x^{2} + x y + y^{2} = 3 x^{2} + 3 \times 11^{2} x + 11^{4}.
$$

Since the left side is divisible by 11, $3 x^{2}$ must also be divisible by 11, so $11 \mid x$. Let $x = 11 z$ (where $z$ is an integer), then $1 = 3 \times 11 z^{2} + 3 \times 11^{2} z + 11^{3}$.
The right side is divisible by 11, but the left side is not, which is a contradiction. Therefore, the equation has no integer solutions in this case.
If $y - x = 11^{3}$, then
$$
1 = x^{2} + x y + y^{2} = 3 x^{2} + 3 \times 11^{3} x + 11^{6}.
$$

In this case, $\Delta < 0$, so the equation has no solutions.
In summary, the solutions to the equation are
$$
\left\{\begin{array}{l}
x = 0, \\
y = 11
\end{array} \text { and } \left\{\begin{array}{l}
x = -11, \\
y = 0.
\end{array}\right.\right.
$$