Solve the following system of equations:

$$
\begin{gathered}
\frac{x-y}{x+z}=a \\
\frac{x^{2}-y^{2}}{x+z}=b \\
\frac{x^{3}+x^{2} y-x y^{2}-y^{3}}{(x+z)^{2}}=\frac{b^{2}}{a^{2} c}
\end{gathered}
$$
(3) Therefore, we must state that $a \neq 0, c \neq 0$. Then, due to (1), $x \neq y$.

Dividing (2) by (1), we obtain

$$
x+y=\frac{b}{a}
$$

The left side of (3) can be transformed using (2) and (4):

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

If $b=0$, then equation (3) is meaningless, from (4) we get $y=-x$, and from (1) we get $2 x=a(x+z)$, which means $z=\frac{2-a}{a} x$. If $b \neq 0$, then from (3) and (5) we get

$$
\frac{1}{x+z}=\frac{1}{a c}, \quad \text { or } \quad x+z=a c
$$

Substituting this value of $(x+z)$ into (1),

$$
x-y=a(x+z)=a^{2} c .
$$

From $\left(1^{\prime}\right)$ and (4),

$$
x=\frac{a^{3} c+b}{2 a}, \quad y=\frac{b-a^{3} c}{2 a}
$$

and thus from (6),

$$
z=a c-x=a c-\frac{a^{3} c+b}{2 a}=\frac{2 a^{2} c-a^{3} c-b}{2 a} .
$$

Miklós Náray (Bp. VIII., Széchenyi g. I. o. t.)