<s>[INST] Equation System with 3 variables "Find all the possible solutions $(x,y,z) \in$  $\Bbb R$ that satisfies the following equation system".
My teacher gave me this problem and i got some clues in how to solve it. I evaluated $1$ and $-1$ and i got these solutions $(1,-1,1)$ and $(-1,-1,-1)$. I think these are the only ones but i can't prove it.
Am i wrong?, are there more solutions?. 
Thanks
$$
\left\{ 
\begin{array}{c}
x+y-z=-1 \\ 
x^2-y^2+z^2=1\\ 
-x^3+y^3+z^3=-1
\end{array}
\right. 
$$
 [/INST] Use the first equation to eliminate $z$: $z=x+y+1$. 
Substituting into the second equation gives
$$x^2-y^2+(x+y+1)^2=1 \Longleftrightarrow (x+y)(x+1)=0\,.$$
Therefore, either $x=-1$ or $x=-y$.
If $x=-1$, then the first equation gives $y=z$, and the third equation becomes $2y^3=-2$. The only solution to this equation is $y=-1$. Hence $(x,y,z)=(-1,-1,-1)$ is a solution. 
If $x=-y$, then the first equation becomes $-z=-1$ so $z=1$. The third equation is now $2y^3=-2$ so $y=-1$ and $x=-y=1$. Hence $(x,y,z)=(1,-1,1)$ is a solution.
There are no other solutions.
</s>