Let's calculate the product $x y z$ if

$$
\begin{gathered}
x+y+z=a \\
x^{2}+y^{2}+z^{2}=b^{2} \\
x^{3}+y^{3}+z^{3}=c^{3}
\end{gathered}
$$
(1) raised to the cube

(4)

$$
(x+y+z)^{3}=x^{3}+y^{3}+z^{3}+6 x y z+3\left(x y^{2}+x^{2} y+x z^{2}+x^{2} z+y z^{2}+y^{2} z\right)=a^{3}
$$

The product of (1) and (2)

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

Subtracting three times (5) from (4) and substituting $c^{3}$ for $x^{3}+y^{3}+z^{3}$ based on (3)

$$
-2 c^{3}+6 x y z=a^{3}-3 a b^{2}
$$

from which

$$
x y z=\frac{a^{3}-3 a b^{2}+2 c^{3}}{6}
$$

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