10.2. Solve the equation $\frac{(2 x-1)^{2}}{x^{3}-2 x^{2}+1}+\frac{4 x-4 x^{2}-1}{x^{3}-3 x^{2}+2}=0$.
Answer: $x_{1}=\frac{1}{2}, x_{2}=-1$.

Solution. Rewrite the equation as $\frac{(2 x-1)^{2}}{x^{3}-2 x^{2}+1}=\frac{(2 x-1)^{2}}{x^{3}-3 x^{2}+2}$. Therefore, either $2 x-1=0$, or $\frac{1}{x^{3}-2 x^{2}+1}=\frac{1}{x^{3}-3 x^{2}+2}$. From the first equation, $x=1 / 2$, and for this value, the denominators of the second equation are not equal to 0, i.e., $x=1 / 2$ is a root. From the second equation, it follows that: $x^{3}-2 x^{2}+1=x^{3}-3 x^{2}+2 \Leftrightarrow$ $x^{2}=1$, i.e., $x= \pm 1$. When $x=1$, the denominator of the first fraction becomes zero, so $x=1$ is not a root. When $x=-1$, both denominators are non-zero.