Solve the following system of equations:

$$
\begin{gathered}
x+y-z=-1 \\
x^{2}-y^{2}+z^{2}=1 \\
-x^{3}+y^{3}+z^{3}=-1
\end{gathered}
$$
Let's rewrite the equations as follows:

$$
\begin{gathered}
x+y=-z-1 \\
x^{2}-y^{2}=1-z^{2} \\
-x^{3}+y^{3}+z^{3}=-1
\end{gathered}
$$

From equation (1), we have \( z = x + y + 1 \). Substituting this into equation (2):

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

Performing the squaring and using the identity \( x^{2}-y^{2}=(x+y)(x-y) \), we rearrange the equation. We get:

$$
2 \cdot (x+1)(x+y)=0
$$

A product is zero if and only if one of its factors is zero.

If

$$
x+1=0, \quad x=-1
$$

## If

then

from equation (1), \(-1+y-z=-1\), which simplifies to \( y = z \). Substituting this into equation (3), we get \( 1 + 2y^{3} = -1 \), from which \( y^{3} = -1 \), and

\( y = -1 \), so the solution to the system of equations is \( x = -1, y = -1, z = -1 \).

Since we performed equivalent transformations throughout, both solutions satisfy the equation.

Of course, we can also verify the correctness of the solutions by substitution, which is useful to ensure that we did not make any calculation errors. The complete solution includes the verification.