6.013. $3\left(x+\frac{1}{x^{2}}\right)-7\left(1+\frac{1}{x}\right)=0$.
## Solution.

Domain of definition: $x \neq 0$.

$$
\begin{aligned}
& 3\left(x+\frac{1}{x^{2}}\right)-7\left(1+\frac{1}{x}\right)=0 \Leftrightarrow \frac{3\left(x^{3}+1\right)-7(x+1) x}{x^{2}}=0 \Leftrightarrow \\
& \Leftrightarrow \frac{3(x+1)\left(x^{2}-x+1\right)-7(x+1) x}{x^{2}}=0 \Leftrightarrow \frac{(x+1)\left(3\left(x^{2}-x+1\right)-7 x\right)}{x^{2}}=0 \Leftrightarrow
\end{aligned}
$$

$\Leftrightarrow \frac{(x+1)\left(3 x^{2}-10 x+3\right)}{x^{2}}=0$.

We have $(x+1)\left(3 x^{2}-10 x+3\right)=0$. From this, $x+1=0, x_{1}=-1$ or $3 x^{2}-10 x+3=0, x_{2}=\frac{1}{3}, x_{3}=3$.

Answer: $x_{1}=-1, x_{2}=\frac{1}{3}, x_{3}=3$.