Solve the following equation:

$$
\sqrt[3]{x}+\sqrt[3]{9-x}=3
$$
I. solution. Let $y=\sqrt[3]{x}$. With this substitution, we can rewrite the equation as follows:

$$
y+\sqrt[3]{9-y^{3}}=3
$$

Subtract $y$ from both sides and cube the equation, then rearrange the equation:

$$
\begin{gathered}
\sqrt[3]{9-y^{3}}=3-y \\
9-y^{3}=27-27 y+9 y^{2}-y^{3} \\
9 y^{2}-27 y+18=0
\end{gathered}
$$

The resulting quadratic equation has two roots: $y_{1}=1$ and $y_{2}=2$. From these, we get $x_{1}=y_{1}^{3}=1$ and $x_{2}=y_{2}^{3}=8$. Substituting these back into the original equation, we can verify that they are indeed solutions.

II. solution. We use the known identity $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$. Cube both sides of the equation:

$$
\begin{gathered}
(\sqrt[3]{x}+\sqrt[3]{9-x})^{3}=3^{3} \\
(\sqrt[3]{x})^{3}+(\sqrt[3]{9-x})^{3}+3 \sqrt[3]{x} \sqrt[3]{9-x}(\sqrt[3]{x}+\sqrt[3]{9-x})=27 \\
9+3 \sqrt[3]{x} \sqrt[3]{9-x}(\sqrt[3]{x}+\sqrt[3]{9-x})=27
\end{gathered}
$$

In the parentheses, we have the left side of the original equation, which we replace with the 3 on the right side of the original equation:

$$
9+3 \sqrt[3]{x} \sqrt[3]{9-x} \cdot 3=27
$$

Rearranging and cubing:

$$
\begin{aligned}
\sqrt[3]{x} \sqrt[3]{9-x} & =2 \\
x(9-x) & =8 \\
0 & =x^{2}-9 x+8
\end{aligned}
$$

The quadratic equation has two roots: $x_{1}=1$ and $x_{2}=8$. Substituting these back into the original equation, we can see that they are indeed solutions.

Remark. In the first solution, we performed only equivalent transformations (cubing the equation, or taking the cube root of both sides), so at the end of the solution, we could also write: "Since we performed only equivalent transformations, 1 and 8 are definitely solutions." However, this is not true in the second solution! In the step where we replaced the expression on the left side of (2) with the right side of the equation, we performed a transformation that is not reversible. (It can only be reversed if we know in advance that the equation is true for some value.)

The fact that such a transformation can indeed result in false roots is shown by the following example:

$$
\sqrt[3]{x}+\sqrt[3]{2-x}=-1
$$

Using the steps of the second solution:

$$
\begin{aligned}
(\sqrt[3]{x}+\sqrt[3]{2-x})^{3} & =-1 \\
(\sqrt[3]{x})^{3}+(\sqrt[3]{2-x})^{3}+3 \sqrt[3]{x} \sqrt[3]{2-x}(\sqrt[3]{x}+\sqrt[3]{2-x}) & =-1 \\
2+3 \sqrt[3]{x}+\sqrt[3]{2-x}(-1) & =-1 \\
\sqrt[3]{x} \sqrt[3]{2-x} & =1 \\
x(2-x) & =1 \\
0 & =x^{2}-2 x+1 \\
x & =1
\end{aligned}
$$

However, this is not a solution to the equation! (It is evident from the outset that the equation has no solution, for example, because the left side is always positive: $\sqrt[3]{x}+\sqrt[3]{2-x}>\sqrt[3]{x}+\sqrt[3]{-x}=0$.)

Those who solved the problem using the method of the second solution but incorrectly stated that they performed only equivalent transformations received 4 points.