
2. Find all triples $(x, y, z)$ of real numbers satisfying

$$
x+y-z=-1, \quad x^{2}-y^{2}+z^{2}=1 \quad \text { and } \quad-x^{3}+y^{3}+z^{3}=-1 \text {. }
$$


2. The first equation yields $z=x+y+1$. Substitution into the second equation gives $x^{2}-y^{2}+$ $(x+y+1)^{2}=1$. Expanding gives $2 x^{2}+2 x y+2 x+2 y=0$, or $2(x+y)(x+1)=0$. We deduce that $x+y=0$ or $x+1=0$. We consider the two cases.

- If $x+y=0$, then $y=-x$ holds. The first equation becomes $z=1$. Substitution into the third equation yields $-x^{3}+(-x)^{3}+1^{3}=-1$, or $x^{3}=1$. We deduce that $(x, y, z)=(1,-1,1)$.
- If $x+1=0$, then $x=-1$ holds. The first equation becomes $z=y$. Substitution into the third equation yields $-(-1)^{3}+y^{3}+y^{3}=-1$, or $y^{3}=-1$. Hence, we have $(x, y, z)=(-1,-1,-1)$.

In total we have found two possible solutions. Substitution into the original equations shows that these are indeed both solutions to the given system.
