6.046. $3 \sqrt[3]{x}-5 \sqrt[3]{x^{-1}}=2 x^{-1}$.
Solution.

From the condition we have $3 \sqrt[3]{x}-\frac{5}{\sqrt[3]{x}}=\frac{2}{x}$.

Let $\sqrt[3]{x}=y, y \neq 0$, and the equation takes the form

$$
3 y-\frac{5}{y}-\frac{2}{y^{3}}=0(y \neq 0) \Leftrightarrow 3 y^{4}-5 y^{2}-2=0 \Leftrightarrow 3\left(y^{2}\right)^{2}-5\left(y^{2}\right)-2=0
$$

from which $y^{2}=2 ; y^{2}=-1$ (not suitable). Then $y_{1,2}= \pm \sqrt{2} \Rightarrow \sqrt[3]{x}=-2$ or $\sqrt[3]{x}=\sqrt{2}, x_{1}=-\sqrt{8}, x_{2}=\sqrt{8}$.

Answer: $x_{1}=-2 \sqrt{2}, x_{2}=2 \sqrt{2}$.