Solve the following equation over the set of real numbers:

$$
1+2 \sqrt{x}-\sqrt[3]{x}-2 \sqrt[6]{x}=0
$$
Among those who solved the task, most arrived at the equation

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

from the original equation by substituting $\sqrt[6]{x}=y$, and noticed that it can be factored as:

$$
(2 y-1)\left(y^{2}-1\right)=0
$$

The product can only be 0 if one of its factors is 0, hence

$$
y=\frac{1}{2}, \quad y= \pm 1
$$

Substituting back, since $\sqrt[6]{x}=-1$ is not possible, we get two solutions for the original equation:

$$
x_{1}=1 \quad \text { and } \quad x_{2}=\frac{1}{64}
$$

It can be verified by substitution that both solutions satisfy the equation.

Remark. Unfortunately, the example was published with a printing error, so many tried to interpret and solve the equation

$$
a+2 \sqrt{x}-\sqrt[3]{x}-2 \sqrt[6]{x}=0
$$

(some considered $a$ as a parameter, others as a variable) and plotted the function for different possible values of $a$, while others tried to determine the roots of the cubic equation with varying success using the Cardano formula.