Example 7. Given that $a \cdot b, c$ are distinct real numbers, and $a b c \neq 0$, solve the system of equations:
$$
\left\{\begin{array}{l}
\frac{x}{b^{3}}-\frac{y}{b^{2}}+\frac{z}{b}=1, \\
\frac{x}{c^{3}}-\frac{y}{c^{2}}+\frac{z}{c}=1, \\
\frac{x}{a^{3}}-\frac{y}{a^{2}}+\frac{z}{a}=1 .
\end{array}\right.
$$
From the form of the original system of equations and the definition of the roots, we know that $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are the three roots of the equation $x t^{3}-y t^{2}+z t=1$. Therefore,
$$
x t^{3}-y t^{2}+z t-1
=x\left(t-\frac{1}{a}\right)\left(t-\frac{1}{b}\right)\left(t-\frac{1}{c}\right),
$$

which means $x t^{3}-y t^{2}+z t-1=x t^{3}-\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) x t^{2}$
$$
+\left(\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}\right) x t-\frac{x}{a b c} \text {. }
$$

By comparing the coefficients of the corresponding terms on both sides of the equation, we get
$$
\left\{\begin{array}{l}
x=a b c, \\
y=a b+b c+c a, \\
z=a+b+c .
\end{array}\right.
$$