6.010. $\frac{1}{x^{3}+2}-\frac{1}{x^{3}+3}=\frac{1}{12}$.
Solution.

Domain of definition: $x \neq-\sqrt[3]{2}, x \neq-\sqrt[3]{3}$.

$\frac{1}{x^{3}+2}-\frac{1}{x^{3}+3}=\frac{1}{12} \Leftrightarrow \frac{x^{6}+5 x^{3}-6}{12\left(x^{3}+2\right)\left(x^{3}+3\right)}=0 \Leftrightarrow$

$\Leftrightarrow\left\{\begin{array}{l}x^{6}+5 x^{3}-6=0, \\ \left(x^{3}+2\right)\left(x^{3}+3\right) \neq 0 .\end{array}\right.$

Let $x^{3}=y$. We get a quadratic equation in terms of $y$: $y^{2}+5 y-6=0$. From this, $y_{1}=-6, y_{2}=1$.

Thus, $x^{3}=-6$ or $x^{3}=1$ and $x_{1}=-\sqrt[3]{6}, x_{2}=1$.

Answer: $x_{1}=-\sqrt[3]{6}, x_{2}=1$.