3. Solve the equation

$$
2 x^{3}=\left(2 x^{2}+x-1\right) \sqrt{x^{2}-x+1} .
$$
Answer: $1 ; \frac{-1-\sqrt{13}}{6}$.

First solution. Let $t=\sqrt{x^{2}-x+1}$. Then $x-1=x^{2}-t^{2}$, and the original equation can be rewritten as $2 x^{3}=\left(3 x^{2}-t^{2}\right) t$. Clearly, $x=0$ is not a root of the original equation. Divide the last equation by $x^{3}$. For the new variable $y=\frac{t}{x}$, we get the equation $y^{3}-3 y+2=0$. Notice that
$y^{3}-3 y+2=(y-1)^{2}(y+2)$. Therefore, $y=1$ or $y=-2$. In the first case, we get $\sqrt{x^{2}-x+1}=x$, from which $x=1$. In the second case, the equation $\sqrt{x^{2}-x+1}=-2 x$ arises, equivalent to the fulfillment of two conditions:

$$
3 x^{2}+x-1=0 ; \quad x \leqslant 0
$$

Therefore, we need to choose the negative root of the obtained quadratic equation.

Second solution. Let $x>0$. Divide both sides of the original equation by $x^{3}$. Since in this case $\sqrt{x^{2}}=x$, we get

$$
2=\left(2+\frac{1}{x}-\frac{1}{x^{2}}\right) \sqrt{1-\frac{1}{x}+\frac{1}{x^{2}}}
$$

The substitution $t=\sqrt{1-\frac{1}{x}+\frac{1}{x^{2}}}$ (note that $t>0$) gives the equation $2=\left(3-t^{2}\right) t$, the only positive root of which is $t=1$. Returning to the variable $x$, we find the positive root of the original equation $x=1$.

Now let $x<0$. The same substitution (with $t>0$) gives the equation $2=-\left(3-t^{2}\right) t$, the only positive root of which is $t=2$. Returning to the variable $x$, we find the negative root of the original equation.

Evaluation. 14 points for a correct solution. If the root $x=1$ is simply guessed, 1 point. If the root $x=1$ is found (as shown in the 2nd solution method) and the negative root is lost (due to the implicit assumption that $x>0$), 4 points.