Solve the following system of equations:

$$
4 x y z=(x+y)(x y+2)=(x+z)(x z+2)=(y+z)(y z+2)
$$
I. solution. Let $b$ be the product $x y z$ and $a$ be the sum $x+y+z$! If $b=0$, then one of $x, y, z$, say $x$, is zero. The equation $4 x y z=(x+y)(x y+2)$ then becomes:

$$
0=2 y, \quad \text { or } \quad y=0
$$

Similarly, we find that $z=0$, so in this case we get the solution $x_{1}=y_{1}=z_{1}=0$. The same solution results from the conditions $y=0$ and $z=0$, but then we use the equations $4 x y z=(y+z)(y z+2)$, or $4 x y z=(z+x)(x z+2)$.

For the rest, we assume that $b \neq 0$, that is, none of $x, y, z$ is zero. We can now write our system of equations as:

$$
\begin{aligned}
4 b & =(a-z)\left(\frac{b}{z}+2\right) \\
4 b & =(a-y)\left(\frac{b}{y}+2\right) \\
4 b & =(a-x)\left(\frac{b}{x}+2\right)
\end{aligned}
$$

Thus, if $a$ and $b$ are already known, $x, y$, and $z$ can only be selected from the solutions of the following equation:

$$
4 b=(a-u)\left(\frac{b}{u}+2\right)
$$

This, for $u \neq 0$, is equivalent to the quadratic equation

$$
2 u^{2}+(5 b-2 a) u-a b=0
$$

whose leading coefficient is not zero. Such an equation can have at most two different solutions, so at least two of $x, y, z$ must be equal.

If $x=y=z$ (and of course they are not zero), then (1) simplifies to

$$
4 x^{3}=2 x\left(x^{2}+2\right)
$$

We can divide by $2 x$, and after rearranging, we get $x^{2}-2=0$, from which we obtain two more solutions:

$$
\begin{array}{r}
x_{2}=y_{2}=z_{2}=\sqrt{2} \\
x_{3}=y_{3}=z_{3}=-\sqrt{2}
\end{array}
$$

There remains the case where two of $x, y, z$ (for example, $x$ and $y$) are equal and the third is different from them. Then both $x$ and $z$ satisfy (3), so the two roots are $x$ and $z$. According to the relationship between the roots and coefficients, $x z=-\frac{a b}{2}$. We know that $b=x y z, a=x+y+z$ and $x=y$, so $b=x^{2} z, a=2 x+z$. Substituting these,

$$
x z=-\frac{x^{2} z(2 x+z)}{2}
$$

We assumed that $x$ and $z$ are not zero, so we can divide by $x z$, and after rearranging,

$$
2 x^{2}+x z+2=0
$$

On the other hand, substituting $x=y$ into the first equation of the system, we get $4 x^{2} z=2 x\left(x^{2}+2\right)$. We can again divide by $2 x$, and after rearranging,

$$
x^{2}-2 x z+2=0
$$

Adding twice (4) to this, we get $5 x^{2}+6=0$, which has no solution in the real numbers. We have thus shown that there is no solution in this case.

Therefore, the only solutions to our system of equations are

$$
x_{1}=y_{1}=z_{1}=0, \quad x_{2}=y_{2}=z_{2}=\sqrt{2}, \quad x_{3}=y_{3}=z_{3}=-\sqrt{2}
$$

It is easy to verify that these three triples indeed satisfy the system of equations.

II. solution: From the equation $(x+y)(x y+2)=(y+z)(y z+2)$, after rearranging, we get

$$
(x-z)\left(y^{2}+x y+y z+2\right)=0
$$

Similarly, we get the equations

$$
(x-y)\left(z^{2}+x z+y z+2\right)=0
$$

$$
(y-z)\left(x^{2}+x y+x z+2\right)=0
$$

If $x, y, z$ were all different, then the expressions in the second parentheses on the left side of the equations would have to be zero, but then their sum would also be zero, whereas their sum is exactly $(x+y+z)^{2}+6$, which is certainly positive. Therefore, at least two of $x, y, z$ must be equal. If, for example, $x=y \neq z$, then the second parenthesis on the left side of (7) is zero, so

$$
0=x^{2}+x y+x z+2=2 x^{2}+x z+2
$$

On the other hand, the equation $4 x y z=(x+y)(x y+2)$ for $x=y$ becomes $4 x^{2} z=2 x\left(x^{2}+2\right)$, or

$$
2 x\left(x^{2}+2-2 x z\right)=0
$$

As we saw in the first solution, if $x=0$, we get the solution

$$
x_{1}=y_{1}=z_{1}=0
$$

If $x \neq 0$, we get the system of equations

$$
\begin{aligned}
& 2 x^{2}+x z+2=0 \\
& x^{2}-2 x z+2=0
\end{aligned}
$$

which, as we saw, has no solution in the real numbers. The only case left is if $x=y=z \neq 0$, in which case we get the solutions

$$
x_{2}=y_{2}=z_{2}=\sqrt{2} \quad \text { and } \quad x_{3}=y_{3}=z_{3}=-\sqrt{2}
$$

Therefore, our system of equations can have three solutions, and these three triples indeed satisfy the system.