Solve the following system of equations:

$$
\begin{gathered}
\frac{x}{a b}+\frac{y}{b c}+\frac{z}{a c}=3 \\
\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=a+b+c \\
c^{2} x+a^{2} y+b^{2} z=a b c(a+b+c) .
\end{gathered}
$$
The given system of equations can also be written as:

$$
\begin{aligned}
\frac{x}{a b}+\frac{y}{b c}+\frac{z}{a c} & =3 \\
\frac{x}{a b} \cdot b+\frac{y}{b c} \cdot c+\frac{z}{a c} \cdot a & =a+b+c \\
\frac{x}{a b} \cdot c+\frac{y}{b c} \cdot a+\frac{z}{a c} \cdot b & =a+b+c
\end{aligned}
$$

Let $\frac{x}{a b}=s, \frac{y}{b c}=u, \frac{z}{a c}=v$, then we get:

$$
\begin{gathered}
s+u+v=3 \\
b s+c u+a v=a+b+c \\
c s+a u+b v=a+b+c
\end{gathered}
$$

From this, $s=1, u=1, v=1$, so $x=a b, y=b c, z=a c$.

(Arthur Paunz, Pécs.)

The problem was also solved by: L. Bánó, J. Dávid, V. Eckstein, M. Fekete, P. Füstös, Z. Harsányi, E. Havas, P. Heimlich, Gy. Jánosy, J. Kiss, B. Lázár, Gy. Meleghy, K. Miklóssy, S. Pichler, E. Rajz, Gy. Schuster, J. Steiger, E. Tandlich, B. Tóth, V. Wáhl.