What integer pairs $(x, y)$ satisfy the equation

$$
x^{3}+y^{3}-3 x^{2}+6 y^{2}+3 x+12 y+6=0
$$
It is easy to recognize that the equation can be rewritten in the following form:

$$
(x-1)^{3}+(y+2)^{3}=1
$$

The sum of two cubes can only be 1 if one is 0 and the other is 1. That is,

$$
\begin{aligned}
& x-1=0, \quad \text { from which } \quad x=1, \quad x-1=1, \quad x=2 \\
& \begin{array}{lll}
\text { or } & \text { or } & \text { that is }
\end{array} \\
& y+2=1, \quad \text { from which } y=-1, \quad y+2=0, \quad y=-2
\end{aligned}
$$

Therefore, two solution pairs satisfy the equation, namely $(1,-1)$ and $(2,-2)$. Since we performed equivalent transformations, these are indeed solutions to the equation.

We still need to prove our statement that if $a, b$ are integers such that $a^{3}+b^{3}=1$, then one of $a$ and $b$ is 0, and the other is 1.

Factoring the product:

$1=a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)$. Since $a$ and $b$ are integers, the product can only be 1 if both factors are either +1 or -1.

For $a+b=1, a=1-b$ and substituting this value into $a^{2}-a b+b^{2}=1$: $(1-b)^{2}-(1-b) b+b^{2}=1$. From this, $b(b-1)=0$. If $b=0$, then $a=1$, and if $b=1$, then $a=0$. If $a+b=-1$, similarly performing the substitution, we arrive at the equation $3 b^{2}+3 b+2=0$, whose discriminant is $D=9-24<0$, so in this case, there is no real solution.