Solve the following equation

$$
\sqrt[3]{\sqrt{5}+x}+\sqrt[3]{\sqrt{5}-x}=\sqrt{5}
$$
I. Solution: Raising both sides of our equation to the third power

$$
\sqrt{5}+x+3 \sqrt{(\sqrt{5}+x)^{2}(\sqrt{5}-x)}+3 \sqrt{(\sqrt{5}+x)(\sqrt{5}-x)^{2}}+\sqrt{5}-x=5 \sqrt{5}
$$

After simplifying and dividing by 3

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

Factoring out the common factor on the left side

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

Dividing this by the original equation

$$
\sqrt[3]{5-x^{2}}=1
$$

from which

$$
5-x^{2}=1
$$

and thus

$$
x_{1}=2, \quad x_{2}=-2
$$

Since we performed identical transformations everywhere (the cube root is unique in the set of real numbers), the roots we found do indeed satisfy the original equation.

Verification:

$$
\begin{aligned}
\sqrt[3]{\sqrt{5}+2} & =\sqrt[3]{\frac{8 \sqrt{5}+16}{8}}=\sqrt[3]{\frac{5 \sqrt{5}+15+3 \sqrt{5}+1}{8}}= \\
& =\sqrt[3]{\left(\frac{\sqrt{5}+1}{2}\right)^{3}}=\frac{\sqrt{5}+1}{2}
\end{aligned}
$$

Similarly

$$
\sqrt[3]{\sqrt{5}-2}=\frac{\sqrt{5}-1}{2}
$$

and thus indeed

$$
\frac{\sqrt{5}+1}{2}+\frac{\sqrt{5}-1}{2}=\sqrt{5}
$$

Mihály Grell (Bp. XVI., Corvin Mátyás g. II. o. t.)

II. Solution: Let

then clearly

$$
u^{3}+v^{3}=2 \sqrt{5}
$$

According to the problem

$$
u+v=\sqrt{5}
$$

Dividing (1) by (2)

$$
u^{2}-u v+v^{2}=2
$$

The square of (2)

$$
u^{2}+2 u v+v^{2}=5
$$

The difference between (4) and (3)

$$
3 u v=3, \quad \text { i.e. } \quad u v=1
$$

and thus

$$
u^{3} v^{3}=(\sqrt{5}+x)(\sqrt{5}-x)=5-x^{2}=1
$$

from which

$$
x= \pm 2
$$