## Task A-1.1.

Determine all triples of real numbers $(x, y, z)$ that satisfy the system of equations

$$
\begin{aligned}
x+y-z= & -1 \\
x^{2}-y^{2}+z^{2}= & 1 \\
-x^{3}+y^{3}+z^{3}= & -1
\end{aligned}
$$
## Solution.

From the first equation, we have $x+y=z-1$. Substituting into the second, we get:

\[
\begin{gathered}
x^{2}-y^{2}=1-z^{2} \\
(x+y)(x-y)=(1-z)(1+z) \\
(z-1)(x-y)=-(z-1)(1+z) \\
(z-1)(x-y+z+1)=0
\end{gathered}
\]

We distinguish two cases: $z=1$ or $x-y+z+1=0$.

If $z=1$, then $x+y=0$, i.e., $x=-y$. Substituting into the third equation, we get $2 y^{3}=-2$, or $y=-1, x=1$.

If $x-y+z+1=0$, then substituting $z=-x+y-1$ into the first equation, we get $2 x=-2$, or $x=-1$.

Now, $z=-(-1)+y-1=y$, so substituting into the third equation, we get $2 y^{3}=-2$, or $y=z=-1$.

Therefore, the solutions to the given system are $(1,-1,1)$ and $(-1,-1,-1)$.