Simplify the following fraction as much as possible:

$$
\frac{(x+y-z)^{3}}{(x+y)(z-x)(z-y)}
$$

given that $z^{3}=x^{3}+y^{3}$.
Our expression only makes sense if

$$
x+y \neq 0, \quad z-x \neq 0, \quad z-y \neq 0
$$

Our expression can also be written as:

$$
\begin{gathered}
\frac{(x+y)^{3}-3(x+y)^{2} z+3(x+y) z^{2}-z^{3}}{(x+y)(z-x)(z-y)}= \\
=\frac{x^{3}+y^{3}+3 x y(x+y)-3(x+y)^{2} z+3(x+y) z^{2}-z^{3}}{(x+y)(z-x)(z-y)}
\end{gathered}
$$

Since according to the condition $x^{3}+y^{3}-z^{3}=0$, our fraction is:

$$
\begin{gathered}
\frac{3 x y(x+y)-3(x+y)^{2} z+3(x+y) z^{2}}{(x+y)(z-x)(z-y)}=\frac{3(x+y)\left(x y-x z-y z+z^{2}\right)}{(x+y)(z-x)(z-y)}= \\
=\frac{3(x+y)(z-x(z-y)}{(x+y)(z-x)(z-y)}
\end{gathered}
$$

Therefore, under the conditions (1),

$$
\frac{(x+y-z)^{3}}{(z+y)(z-x)(x-y)}=3
$$

if $z^{3}=x^{3}+y^{3}$.

Péch Antal (Bp., VIII., Piarist g. II. o. t.)