## Task 1 - 261231

Determine all triples $(x ; y ; z)$ of real numbers that satisfy the following system of equations (1), (2), (3):

$$
\begin{aligned}
x+y-z & =1 \\
x^{2}-y^{2}+z^{2} & =1 \\
-x^{3}+y^{3}+z^{3} & =-1
\end{aligned}
$$
By substituting $z=x+y-1$ from equation (1) into (2), we obtain

$$
x^{2}-y^{2}+z^{2}-1=x^{2}-y^{2}+(x+y)^{2}-2(x+y)=2(x+y)(x-1)=0
$$

Therefore, the following case distinction is appropriate:

1. Case: $x+y=0$ or $y=-x$.

From (1), it immediately follows that $z=-1$, and from (3) we then get $2 x^{3}=0$, hence $x=y=0$.

2. Case: $x=1$.

Substituting into (1) then leads to $y=z$, and from (3) we again get $2 y^{3}=0$, hence $y=z=0$.

In total, we must have

$$
(x, y, z) \in\{(0,0,-1),(1,0,0)\}
$$

and these are indeed solutions, as can be verified immediately.