Solve the following system of equations:

$$
\begin{aligned}
a^{3}+a^{2} x+a y+z & =0 \\
b^{3}+b^{2} x+b y+z & =0 \\
c^{3}+c^{2} x+c y+z & =0
\end{aligned}
$$
I. Solution: First of all, it is obvious that among the coefficients $a, b, c$, there cannot be two equal, otherwise we would not have 3 independent equations. Subtracting (2) from (1),

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

that is, dividing by $(a-b)$,

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

Similarly, subtracting (3) from (1) and dividing by $(a-c)$,

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

Subtracting (5) from (4),

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

from which

$$
a+b+c+x=0
$$

and thus

$$
x=-(a+b+c)
$$

Substituting this value of $x$ into (4),

$$
\begin{gathered}
y=-a^{2}-a b-b^{2}+(a+b)(a+b+c)= \\
=-a^{2}-a b-b^{2}+a^{2}+2 a b+b^{2}+a c+b c=a b+a c+b c
\end{gathered}
$$

Substituting the values of $x$ and $y$ into (1),

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

Deseő Zoltán (Bp. X., I. László g. II. o. t.)

II. Solution: Consider $x, y, z$ as the coefficients of the cubic equation in $v$,

$$
v^{3}+x v^{2}+y v+z=0
$$

whose roots, according to the problem, are $-v_{1}=a, v_{2}=b, v_{3}=c$.

The factorized form is thus

$$
(v-a)(v-b)(v-c)=0
$$

or

$$
v^{3}-(a+b+c) v^{2}+(a b+a c+b c) v-a b c=0
$$

From this, it is clear that

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

Kovács László (Debrecen, Ref. g. II. o. t.)