## Task $9 / 89$

Determine all non-negative real numbers $x$ that satisfy the equation:

$$
\sqrt[3]{x}+\sqrt[4]{x}=6 \sqrt[6]{x}
$$
One quickly recognizes that $x=0$ is a solution of the equation. For the following, one can therefore assume $x \neq 0$. Dividing the equation by $\sqrt[6]{x}$, and applying the laws of root arithmetic, yields

$$
\sqrt[12]{x^{2}}+\sqrt[12]{x}=6
$$

This quadratic equation in $\sqrt[12]{x}$ has the solutions $\sqrt[12]{x}=-0.5 \pm 2.5$. Since $\sqrt[12]{x}>0$, it follows that $\sqrt[12]{x}=2$, and thus $x=2^{12}=4096$. Therefore, there are two solutions: $x_{1}=0, x_{2}=4096$.