Solve the following equation:

$$
\sqrt[4]{x}+\sqrt{x}=2 \sqrt[3]{x}
$$
Since a negative number cannot appear under the square root, $x$ cannot be negative either. Therefore, there exists a positive $y$ such that $y^{12}=x$. For this $y$, we get the following equation from the original:

$$
y^{3}+y^{6}=2 \cdot y^{4}
$$

which, with the constraint that $y$ is non-negative, is equivalent to the original. $y=0$ is possible. This leads to the solution $y_{1}=0$, which means $x_{1}=0$; and by dividing by $y^{3}$, we get the equation

$$
y^{3}-2 y+1=0
$$

This equation has the root 1, from which we get $y_{2}=1$, which means $x_{2}=1$ is also a solution. Now we can divide the equation under (1) by $(y-1)$ and thus arrive at the equation

$$
y^{2}+y-1=0
$$

Solving this for $y$ yields two roots:

$$
y_{3}=(-1+\sqrt{5}) / 2 \quad \text { and } \quad y_{4}=(-1-\sqrt{5}) / 2
$$

Since $y$ cannot be negative, $y_{4}$ is not a valid solution.

Therefore, the solutions to the original equation are: $x_{1}=0, x_{2}=1$ and $x_{3}=[(-1+\sqrt{5}) / 2]$.