6.063. $\frac{\sqrt{x}+\sqrt[3]{x}}{\sqrt{x}-\sqrt[3]{x}}=3$.
## Solution.

Domain of definition: $0<x \neq 1$.

Rewrite the equation as

$$
\frac{\sqrt[6]{x^{3}}+\sqrt[6]{x^{2}}}{\sqrt[6]{x^{3}}-\sqrt[6]{x^{2}}}=3 \Leftrightarrow \frac{\sqrt[6]{x^{3}}(\sqrt[6]{x}+1)}{\sqrt[6]{x^{3}}(\sqrt[6]{x}-1)}=3 \Leftrightarrow \frac{\sqrt[6]{x}+1}{\sqrt[6]{x}-1}=3 \Leftrightarrow
$$

$$
\begin{aligned}
& \Leftrightarrow \frac{\sqrt[6]{x}+1}{\sqrt[6]{x}-1}-3=0, x \neq 1 \Leftrightarrow \frac{\sqrt[6]{x}+1-3 \sqrt[6]{x}+3}{\sqrt[6]{x}-1}=0 \Leftrightarrow \\
& \Leftrightarrow-2 \sqrt[6]{x}+4=0 \Leftrightarrow \sqrt[6]{x}=2, x=2^{6}=64
\end{aligned}
$$

Answer: $x=64$.