Example 31 (53rd Romanian Mathematical Olympiad) Find the complex solutions of the system of equations $\left\{\begin{array}{l}x(x-y)(x-z)=3, \\ y(y-x)(y-z)=3, \\ z(z-x)(z-y)=3\end{array}\right.$.
Due to the symmetry of the letters $x, y, z$, this prompts us to explore symmetry.
It is easy to know that $x \neq 0, y \neq 0, z \neq 0, x \neq y, y \neq z, z \neq x$.
By dividing the first two equations of the system, we get $x^{2}+y^{2}=y z+z x$.
Similarly, we get $y^{2}+z^{2}=x y+z x, z^{2}+x^{2}=x y+y z$.
Adding the above three equations, we get $x^{2}+y^{2}+z^{2}=x y+y z+z x$.
Subtracting the first two of the above three equations, we get $x^{2}-z^{2}=y z-x y$, i.e., $x(x+y)=z(y+z)$, thus $x+y+z=0$.
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 (1), we get $z^{2}=x y$, i.e., $z^{3}=x y z$.
Similarly, we get $x^{3}=y^{3}=x y z$.
Let $x y z=a$, since $x, y, z$ are distinct, we set $x=\sqrt[3]{a}, y=\sqrt[3]{a} \cdot \omega, z=\sqrt[3]{a} \cdot \omega^{2}$, where $\omega=\frac{-1+\sqrt{3} \mathrm{i}}{2}$.
Substituting them into each equation of the original system, we get $a(1-\omega)\left(1-\omega^{2}\right)=3$, i.e., $a \cdot\left(2-\omega-\omega^{2}\right)=a \cdot 3=3$, hence $a=1$.
Therefore, the solution $(x, y, z)$ of the original system of equations is a permutation of $\left\{1, \omega, \omega^{2}\right\}$.