## Task B-2.1.

Solve the system of equations

$$
\begin{gathered}
x+y-z=2 \\
x^{2}+y^{2}-z^{2}=-4 \\
x^{3}+y^{3}-z^{3}=8
\end{gathered}
$$
## Solution.

The first equation of the system can be written in the form $x+y=z+2$. By squaring, we get $x^{2}+y^{2}+2xy=z^{2}+4z+4$, and by subtracting the second given equation from this, we obtain $2xy=4z+8$, or $xy=2z+4$.

The sum of cubes $x^{3}+y^{3}$ in the third equation of the system can be written in a more convenient form as $(x+y)^{3}-3xy(x+y)-z^{3}=20$, and we substitute $x+y=z+2$ and $xy=2z+4$.

We get an equation with one unknown: $(z+2)^{3}-3(2z+4)(z+2)-z^{3}=8$, whose solution is $z=-2$.

Therefore, we have

$$
x+y=0 \quad xy=0
$$

The only solution to this system is $(0,0)$. Therefore, the solution to the given system is $(x, y, z)=(0,0,-2)$.