9. Solve the equation

$$
\sqrt[3]{x^{2}+2 x}+\sqrt[3]{3 x^{2}+6 x-4}=\sqrt[3]{x^{2}+2 x-4}
$$

Solution: Let's make the substitution $t=x^{2}+2 x$, as a result of which the equation takes the form $\sqrt[3]{t}+\sqrt[3]{3 t-4}=\sqrt[3]{t-4}$. Further, we have

$$
\begin{aligned}
& \sqrt[3]{t}+\sqrt[3]{3 t-4}=\sqrt[3]{t-4} \Longleftrightarrow \\
& \Longleftrightarrow t+3 t-4+3 \sqrt[3]{t(3 t-4)}(\sqrt[3]{t}+\sqrt[3]{3 t-4})=t-4 \Longleftrightarrow \\
& \Longleftrightarrow \sqrt[3]{t(3 t-4)} \sqrt[3]{t-4}=-t \Longleftrightarrow t(3 t-4)(t-4)=-t^{3} \Longleftrightarrow \\
& \Longleftrightarrow t(t-2)^{2}=0, \text { from which } t=0 \text { or } t=2 .
\end{aligned}
$$

Thus, $x^{2}+2 x=0$, from which $x=0 ;-2$, or $x^{2}+2 x-2=0$, from which $x=-1 \pm \sqrt{3}$.

Answer: $\{-2 ; 0 ;-1-\sqrt{3} ;-1+\sqrt{3}\}$.

B. Solve problems $10-12$ in as many ways as possible. Different methods are considered those that use different mathematical ideas, as well as different technical approaches to implementing the same idea. Indicate the place of each of the methods used in the school mathematics course.
2.I.9. Comment. The following error was made in the given solution. The equation $\sqrt[3]{t(3 t-4)} \cdot \sqrt[3]{t-4}=-t$ is not equivalent to the original equation, but is only a consequence of it, so its roots may not be roots of the original equation. Indeed, the root $t=2$ is not a root of the original equation. The correct answer is $\{-2 ; 0\}$.

The idea of the given reasoning was to transition from the equation $a+b=d$ to the equation $a^{3}+b^{3}+3 a b d=d^{3}$, which is a consequence of the first equation. As shown, the roots of the second equation are not necessarily roots of the first equation. However, let's determine what the second equation is equivalent to. For convenience, let $c=-d$. Then we obtain the equation

$$
a^{3}+b^{3}+c^{3}-3 a b c=0
$$

It is well known that the following factorization holds:

$$
a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-a c\right)
$$

Since

$$
2\left(a^{2}+b^{2}+c^{2}-a b-b c-a c\right)=(a-b)^{2}+(b-c)^{2}+(c-a)^{2}
$$

then

$$
a^{3}+b^{3}+c^{3}-3 a b c=0 \Longleftrightarrow a+b+c=0 \text { or } a=b=c .
$$

In the given solution, the "extraneous" root turned out to be the value of $t$ such that $t=3 t-4=4-t$, i.e., $t=2$.