Solve the following system of equations:

$$
a^{2} b^{2}-a^{2}-a b+1=0 \\
a^{2} c-a b-a-c=0 \\
a b c=-1 .
$$
From equation (3), solve for $c$ and substitute it into equation (2):

$$
a^{2} \cdot\left(-\frac{1}{a b}\right)-a b-a-\left(-\frac{1}{a b}\right)=0
$$

Rearranging the equation, we get

$$
1-a^{2}-a^{2} b^{2}-a^{2} b=0
$$

Subtract equation (1) from equation (4):

$$
2 a^{2} b^{2}+a^{2} b-a b=0
$$

Factoring the equation, we get $a b(2 a b+a-1)=0$. A product is zero if and only if one of its factors is zero. From equation (3), $a b \neq 0$, so $2 a b+a-1=0$, from which

$$
a=\frac{1}{2 b+1}, \quad\left(2 b+1 \neq 0, \quad b \neq-\frac{1}{2}\right)
$$

Substituting this into equation (1), we get

$$
\left(\frac{1}{2 b+1}\right)^{2} b^{2}-\left(\frac{1}{2 b+1}\right)^{2}-\frac{1}{2 b+1} \cdot b+1=0
$$

This results in a quadratic equation in $b$, from which, after rearrangement, $3 b^{2}+3 b=0$ is obtained, or $3 b(b+1)=0$. Since $b \neq 0$, $b=-1$.

From $a=\frac{1}{2 b+1}$, we get $a=-1$, and from (3), $c=-1$.

The solution to the system of equations is: $a=-1, b=-1, c=-1$.

We can verify by substitution that these values indeed satisfy the system of equations.