Solve the following equation

$$
\sqrt[3]{\sqrt{5}+x}+\sqrt[3]{\sqrt{5}-x}=\sqrt[3]{5 \sqrt{5}}
$$
$$
\sqrt[3]{\sqrt{5}+x}+\sqrt[3]{\sqrt{5}-x}=\sqrt[3]{5 \sqrt{5}}
$$

Cubing both sides of the equation, we get:

$$
\begin{gathered}
\sqrt{5}+x+3 \sqrt[3]{(\sqrt{5}+x)^{2}(\sqrt{5}-x)}+ \\
+3 \sqrt[3]{(\sqrt{5}+x)(\sqrt{5}-x)^{2}}+\sqrt{5}-x=5 \sqrt{5} \\
3 \sqrt[3]{\left(5-x^{2}\right)}(\sqrt[3]{\sqrt{5}+x}+\sqrt[3]{\sqrt{5}-x})=3 \sqrt{5}
\end{gathered}
$$

Substituting the value of the second factor from (1):

$$
\sqrt[3]{\left(5-x^{2}\right) 5 \sqrt{5}}=\sqrt{5}
$$

and cubing:

$$
\left(5-x^{2}\right) 5 \sqrt{5}=\sqrt{5}
$$

we finally get the equation

$$
\begin{gathered}
5-x^{2}=1 \\
x^{2}=4 \\
x= \pm 2
\end{gathered}
$$

(Gustáv Szabó and Jenő Schiller, 6th grade students, Győr.)

The problem was also solved by: 7th grade students from the Debrecen main real school; Tivadar Heymann, Győr; Arthur Herusch, Pozsony; György Jankovich and Sándor Kugel, Losoncz; Győző Prónay, Beszterczebánya; Mihály Seidner, Losoncz. Miklós Malesevits, main gymnasium teacher, Zombor.