3. Solve the system of equations in the set of complex numbers

$$
\begin{aligned}
& x(x-y)(x-z)=3 \\
& y(y-x)(y-z)=3 \\
& z(z-x)(z-y)=3
\end{aligned}
$$
3. Note that the numbers $x, y$ and $z$ must be mutually distinct.

From the first and second equations, we get

$$
x(x-z)=y(z-y) \quad \text { i.e. } \quad x^{2}+y^{2}=z(x+y)
$$

Similarly, we obtain $y^{2}+z^{2}=x(y+z)$ and $z^{2}+x^{2}=y(z+x)$.

5 points

From these three equations, it follows that

$$
x^{2}+y^{2}+z^{2}=x y+y z+z x \quad \text { and } \quad x^{2}=y z, y^{2}=z x, z^{2}=x y . \quad 5 \text { points }
$$

From here, $x^{2}-y^{2}=y z-z x$, i.e., $(x+y)(x-y)=-z(x-y)$, from which we get $x+y+z=0$.

Now from the first equation, we get

$$
x\left(x^{2}-(y+z) x+y z\right)=3 \quad \Leftrightarrow \quad x\left(x^{2}-(-x) x+x^{2}\right)=3 \quad \Leftrightarrow \quad x^{3}-1=0.5 \text { points }
$$

By factorization, we get

$$
(x-1)\left(x^{2}+x+1\right)=0
$$

The solutions to this equation are from the set $\left\{1, \frac{-1+i \sqrt{3}}{2}, \frac{-1-i \sqrt{3}}{2}\right\}$. Therefore, $(x, y, z)=\left(1, \frac{-1+i \sqrt{3}}{2}, \frac{-1-i \sqrt{3}}{2}\right)$, or any permutation of these numbers.