Solve the following system of equations:

$$
\begin{gathered}
x+\frac{1}{x}+y+\frac{1}{y}=a \\
x^{3}+\frac{1}{x^{3}}+y^{3}+\frac{1}{y^{3}}=b
\end{gathered}
$$
Given

$$
x+\frac{1}{x}=u \quad \text { and } \quad y+\frac{1}{y}=v
$$

then

$$
x^{3}+\frac{1}{x^{3}}=u^{3}-3 u \quad \text { and } \quad y^{3}+\frac{1}{y^{3}}=v^{3}-3 v
$$

Thus, our equations become:

$$
u+v=a
$$

$$
u^{3}+v^{3}=b+3 a
$$

If we subtract (4) from the cube of (3), we get

$$
3 u^{2} v+3 u v^{2}=a^{3}-b-3 a
$$

from which

$$
u v=\frac{a^{3}-3 a-b}{3 a}
$$

We thus know $(u+v)$ and $u v$. Therefore, $u$ and $v$ are the roots of the following equation:

$$
3 a z^{2}-3 a^{2} z+a^{3}-3 a-b=0
$$

If the roots of this equation are $z_{1}=u$ and $z_{2}=v$, then

$$
x+\frac{1}{x}=z_{1} \text { and } y+\frac{1}{y}=z_{2}
$$

from which the values of $x$ and $y$ can be determined.

(Géza Sasvári, Pécs.)

The problem was also solved by: Bayer B., Bertrám J., Faith F., Filkorn J., Freibauer E., Kerekes T., Krausz B., Krisztián Gy., Kürth A., Lukhaub Gy., Lupsa Gy., Messik G., Messik V., Oltay K., Perl Gy., Póka Gy., Sasvári J., Scharff J.