Solve the following equation:

$$
\frac{3 x^{3}}{x^{3}-1}-\frac{x}{x-1}=2
$$
Solution. Due to the denominators: $x \neq 1$.

We know that $(x-1)\left(x^{2}+x+1\right)=x^{3}-1$, therefore $\frac{3 x^{3}-x\left(x^{2}+x+1\right)}{x^{3}-1}=2$.

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

thus $x=1, x=-2$. Due to the constraint, the only solution is: $x=-2$.