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

$$
x+y-z=-1, \quad x^{2}-y^{2}+z^{2}=1 \quad \text { and } \quad -x^{3}+y^{3}+z^{3}=-1 \text {. }
$$
2. The first equation gives $z=x+y+1$. We substitute this into the second equation, so we find: $x^{2}-y^{2}+(x+y+1)^{2}=1$. Expanding gives $2 x^{2}+2 x y+2 x+2 y=0$, or equivalently $2(x+y)(x+1)=0$. We see now that $x+y=0$ or $x+1=0$. We consider the two cases separately.

- If $x+y=0$, then $y=-x$. The first equation gives $z=1$. Substituting into the third equation gives $-x^{3}+(-x)^{3}+1^{3}=-1$, or $x^{3}=1$. We see that $(x, y, z)=(1,-1,1)$.
- If $x+1=0$, then $x=-1$. The first equation gives $z=y$. Substituting into the third equation gives $-(-1)^{3}+y^{3}+y^{3}=-1$, or $y^{3}=-1$. Thus $(x, y, z)=(-1,-1,-1)$.

In total, we find two possible solutions. Substituting back into the original equations shows that these are indeed solutions.