For what values of the parameter $p$ will the equation

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

have exactly one solution?
Let the function on the left side of the equation be denoted by $f$. It can be seen that for all real $x$, $f(x) = f(2-x)$, so if $x_0$ is a solution, then $(2-x_0)$ is also a solution. If $x_0 \neq 1$, then $x_0 \neq 2-x_0$. The $x_0 = 1$ is a solution only when $p = 2$, so if $p \neq 2$, then (1) has an even number of solutions - possibly zero.

In the case of $p = 2$ - when $x = 1$ is a solution - we still need to check if the equation has any other solutions.

Using the identity $(u+v)^3 = u^3 + v^3 + 3uv(u+v)$, we get by cubing:

$$
x + (2-x) + 3 \cdot \sqrt[3]{x(2-x)}(\sqrt[3]{x} + \sqrt[3]{2-x}) = 8
$$

Notice that if $x_0$ is a solution to (1), which is equivalent to (2), that is, $\sqrt[3]{x_0} + \sqrt[3]{2-x_0} = 2$, then for $x_0$ the following must hold from (2) after rearrangement:

$$
\begin{gathered}
2 + 3 \cdot \sqrt[3]{x_0(2-x_0)} \cdot 2 = 8, \quad \text{that is} \\
\sqrt[3]{x_0(2-x_0)} = 1
\end{gathered}
$$

Cubing and rearranging again, we get $(x_0 - 1)^2 = 0$, so in the case of $p = 2$, equation (1) indeed has no roots other than 1, hence the answer to the problem is: $p = 2$.

Remark. We show that if $x \neq 1$, then $0 < \sqrt[3]{x} + \sqrt[3]{2-x} < 2$. The inequality $\sqrt[3]{x} + \sqrt[3]{2-x} > 0$ immediately follows from cubing the equivalent inequality $\sqrt[3]{x} > \sqrt[3]{x-2}$.

According to the identity already cited,

$$
\begin{gathered}
2 = (\sqrt[3]{x})^3 + (\sqrt[3]{2-x})^3 = (\sqrt[3]{x} + \sqrt[3]{2-x})^3 - 3 \cdot \sqrt[3]{2-x} \cdot (\sqrt[3]{x} + \sqrt[3]{2-x}) = \\
= f^3(x) - 3 \cdot \sqrt[3]{x(2-x)} \cdot f(x)
\end{gathered}
$$

The coefficient of $f(x)$ on the right side of (3), $\sqrt[3]{x(2-x)}$, is less than or equal to 1, since cubing the inequality $(x-1)^2 \geq 0$ gives this, and it is also clear that if $x \neq 1$, then equality cannot hold.

We have obtained that if $x \neq 1$, then $2 > f^3(x) - 3f(x)$, and from this, after rearrangement and factoring,

$$
0 > f^3(x) - 3f(x) - 2 = (f(x) - 2)(f(x) + 1)^2
$$

The second factor on the right side of the product, $(f(x) + 1)^2$, is positive because $f(x) > 0$, so the product itself can only be negative if the first factor, $f(x) - 2$, is negative, and this is precisely what we wanted to prove.