Solve the following equation on the set of real numbers:

$$
\sqrt[5]{x^{3}+2 x}=\sqrt[3]{x^{5}-2 x}
$$
Solution. A function appears on each side of the equation.

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

If for some $x_{0}$, $f\left(x_{0}\right)=g\left(x_{0}\right)$, then let $t_{0}$ denote this common function value. Then

$$
\left(f\left(x_{0}\right)\right)^{5}=x_{0}^{3}+2 x_{0}=t_{0}^{5}, \quad\left(g\left(x_{0}\right)\right)^{3}=x_{0}^{5}-2 x_{0}=t_{0}^{3}
$$

From the sum of the two expressions,

$$
x_{0}^{5}+x_{0}^{3}=t_{0}^{5}+t_{0}^{3}
$$

The function $x^{5}+x^{3}$ is strictly increasing over the entire real domain, so this can only be true if $x_{0}=t_{0}$. Therefore,

$$
x_{0}^{3}+2 x_{0}=x_{0}^{5}, \quad \text { and } \quad x_{0}^{5}-2 x_{0}=x_{0}^{3}
$$

Both equations express the same thing: $x_{0}^{5}-x_{0}^{3}-2 x_{0}=0$. This yields $x_{0}\left(x_{0}^{4}-x_{0}^{2}-2\right)=0$, whose real solutions are 0, and $\pm \sqrt{\frac{1 \pm \sqrt{1+8}}{2}}$, which are $\sqrt{2}$ and $-\sqrt{2}$.

These three numbers indeed satisfy the original equation, and there can be no other real solution.