Example 5. Given that $a, b, c$ are distinct real numbers, and $a b c \neq 0$.
Solve the system of equations $\left\{\begin{array}{l}\frac{x}{b^{3}}-\frac{y}{b^{2}}+\frac{z}{b}=1, \\ \frac{x}{c^{3}}-\frac{y}{c^{2}}+\frac{z}{c}=1, \\ \frac{x}{a^{3}}-\frac{y}{a^{2}}+\frac{z}{a}=1 .\end{array}\right.$
(1987, Fujian Province Mathematics Competition)
Solving this problem directly is quite challenging. From each equation in the system, we can see that the structure is the same. Here, we might as well treat $x, y, z$ as known quantities and consider $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ as the unknowns. It is easy to see that $\frac{1}{a}$, $\frac{1}{b}$, $\frac{1}{c}$ are the three roots of the equation $x t^{3}-y t^{2}+z t-1=0$.
By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{y}{x}, \\
\frac{1}{a} \cdot \frac{1}{b}+\frac{1}{a} \cdot \frac{1}{c}+\frac{1}{b} \cdot \frac{1}{c}=\frac{z}{x}, \\
\frac{1}{a} \cdot \frac{1}{b} \cdot \frac{1}{c}=\frac{1}{x} .
\end{array}\right.
$$

From (3), we get $x=a b c$.
Substituting (3) into (1), we get $y=a b+b c+c a$.
Substituting (3) into (2), we get $z=a+b+c$.
$$
\therefore\left\{\begin{array}{l}
x=a b c, \\
y=a b+b c+c a, \\
z=a+b+c .
\end{array}\right.
$$