Solve the following system of equations:

$$
\begin{array}{r}
a^{3}+a^{2} x+a y+z=0 \\
b^{3}+b^{2} x+b y+z=0 \\
c^{3}+c^{2} x+c y+z=0
\end{array}
$$
From the first and second equations:

$$
a^{3}-b^{3}+\left(a^{2}-b^{2}\right) x+(a-b) y=0
$$

or

$$
a^{2}+a b+b^{2}+(a+b) x+y=0
$$

Similarly, from the first and third equations

$$
a^{2}+a c+c^{2}+(a+c) x+y=0
$$

From (1) and (2)

$$
\begin{aligned}
& x=-(a+b+c) \\
& y=a b+b c+c a
\end{aligned}
$$

thus

$$
z=-a^{3}-a^{2} x-a y=-a b c
$$