5. Find the complex solutions of the system of equations $\left\{\begin{array}{l}x(x-y)(x-z)=3, \\ y(y-x)(y-z)=3 \\ z(z-x)(z-y)=3\end{array}\right.$.
(53rd Romanian Mathematical Olympiad (Second Round))
Solution: It is easy to know that $x \neq 0, y \neq 0, z \neq 0, x \neq y, y \neq z, z \neq x$. From the first two equations, we can get $x^{2}+y^{2}=y z+z x$.
Similarly, we can get $y^{2}+z^{2}=x y+z x, z^{2}+x^{2}=x y+y z$.
Adding the above three equations, we get
$$
x^{2}+y^{2}+z^{2}=x y+y z+z x \text {. }
$$

Subtracting the first two equations, we get $x^{2}-z^{2}=y z-x y$, which means $x+y+z=0$.
From $(x+y+z)^{2}=0$, we get $x y+y z+z x=0$.
Substituting $x^{2}+y^{2}=y z+z x$ into equation (1), we get $z^{2}=x y$, i.e., $z^{3}=x y z$.
Similarly, we can get $x^{3}=y^{3}=x y z$.
Let $x y z=a$, since $x, y, z$ are distinct, we set $x=\sqrt[3]{a}$, $y=\sqrt[3]{a}, z=\sqrt[3]{a} \varepsilon^{2}$, where $\varepsilon=\frac{-1+\sqrt{3} i}{2}$. Substituting them into the original equation, we get $a(1-\varepsilon)\left(1-\varepsilon^{2}\right)=3$. Thus, we have $a=1$. Therefore, the solution $(x, y, z)$ of the original system of equations is a permutation of $\left\{1, \varepsilon, \varepsilon^{2}\right\}$.