Solve the following equation:

$$
\sqrt[3]{x}+\sqrt[3]{1-x}=\frac{3}{2}
$$
By cubing both sides of the equation and using the identity $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$, we get

$$
\sqrt[3]{x(1-x)}(\sqrt[3]{x}+\sqrt[3]{1-x})=\frac{19}{24}
$$

Here, according to the original equation, $\sqrt[3]{x}+\sqrt[3]{1-x}=\frac{3}{2}$. Substituting this into (2), we get

$$
\frac{3}{2} \sqrt[3]{x(1-x)}=\frac{19}{24}
$$

From this, after another cubing and rearrangement, we obtain the equation

$$
x^{2}-x+\left(\frac{19}{36}\right)^{3}=0
$$

The roots of this equation are:

$$
\begin{aligned}
& x_{1}=\frac{1}{2}+\frac{31 \sqrt{5}}{216} \approx 0.8209 \\
& x_{2}=\frac{1}{2}-\frac{31 \sqrt{5}}{216} \approx 0.1791
\end{aligned}
$$

It is clear that the roots of (1) are also roots of (4), so if (1) has roots, they can only be $x_{1}$ or $x_{2}$—or possibly both. However, based on the steps so far, we cannot yet conclude that $x_{1}$ and $x_{2}$ are roots of (1), since we substituted (1) into (2). Thus, the fact that, for example, $x_{1}$ is a root of (4), and thus of the equivalent (3), would imply that $x_{1}$ is a root of (2), and thus of (1), only if we already knew that $\sqrt[3]{x_{1}}+\sqrt[3]{1-x_{1}}=\frac{3}{2}$. The reverse implication would only follow from the fact that $x_{1}$ is a root of (1) if we already knew that $x_{1}$ is a root of (1). This is a circular argument.

Recognizing that $x_{1}=\left(\frac{9+\sqrt{5}}{12}\right)^{3}$ and $x_{2}=1-x_{1}=\left(\frac{9-\sqrt{5}}{12}\right)^{3}$, the verification of the roots can be done by substitution:

$$
\sqrt[3]{x_{1}}+\sqrt[3]{1-x_{1}}=\sqrt[3]{x_{2}}+\sqrt[3]{1-x_{2}}=\sqrt[3]{x_{1}}+\sqrt[3]{x_{2}}=\frac{9+\sqrt{5}}{12}+\frac{9-\sqrt{5}}{12}=\frac{3}{2}
$$

Thus, the roots of the equation are: $\frac{1}{2}+\frac{31 \sqrt{5}}{216}$ and $\frac{1}{2}-\frac{31 \sqrt{5}}{216}$.

Remarks. 1. Many incomplete solutions were submitted for the problem. Almost every solver considered the substitution of $\sqrt[3]{x}+\sqrt[3]{1-x}$ as an equivalent transformation. Another common mistake occurred during the verification of the roots. Many used approximate values, but even if their result approximated $\frac{3}{2}$ to several significant digits, nothing can be concluded from this.

2. The following example shows that the "critical" transformation is not necessarily equivalent. It is not just a theoretical error that could lead to a mistake! If we start from the equation

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

then the method followed in the solution now leads to the quadratic equation

$$
4 x^{2}-4 x+1=0
$$

The root of this equation is $x=\frac{1}{2}$, but this is clearly not a root of (1'), which has no roots. (F. T.)