3-ча 1. Solve the system of equations:

$$
\left\{\begin{aligned}
3 x y z - x^{3} - y^{3} - z^{3} & = b^{3} \\
x + y + z & = 2 b \\
x^{2} + y^{2} - z^{2} & = b^{2}
\end{aligned}\right.
$$
Consider first the case when $b=0$. In this case, the last two equations can be written as $z=-x-y$ and $z^{2}=x^{2}+y^{2}$. Squaring the first of these, we get $x y=0$. Therefore, $x=0, z=-y$ or $y=0, z=-x$. The first equation of the original system is satisfied in these cases.

Now consider the case when $b \neq 0$. We will use the identity

$$
3 x y z-x^{3}-y^{3}-z^{3}=(x+y+z)\left(x y+y z+x z-x^{2}-y^{2}-z^{2}\right)
$$

From the first and second equations, it follows that $x y+y z+x z-x^{2}-y^{2}-z^{2}=\frac{b^{2}}{2}$. Squaring the equation $x+y+z=2 b$, we get $x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 x z=4 b^{2}$. Therefore, $x^{2}+y^{2}+z^{2}=b^{2}$ and $x y+y z+x z=\frac{3}{2} b^{2}$. Comparing the first of these equations with the last equation of the original system, we obtain $z=0$. Thus, $x^{2}+y^{2}=b^{2}$ and $x y=\frac{3}{2} b^{2}$. Solving this system of equations, we find $x=\left(1 \pm \sqrt{\frac{-1}{2}}\right) b$, $y=\left(1 \mp \sqrt{\frac{-1}{2}}\right) b$.