Let's determine the integer solutions of the equation $\sqrt[3]{a}+\sqrt[3]{b}=\sqrt[3]{c}$.
When $c=0$, $b=-a$ provides all solutions. In the case of $a=0$ (or $b=0$), $c=b$ (or $c=a$) gives the solutions. Otherwise, it can be assumed that none of $a, b, c$ are 0. Thus, we can divide by $\sqrt[3]{c}$; and in this case, we arrive at the equation $\sqrt[3]{u}+\sqrt[3]{v}=1$, where $u=\frac{a}{c}$ and $v=\frac{b}{c}$ are rational numbers. According to the relationship

$$
u+v=(\sqrt[3]{u}+\sqrt[3]{v})\left(\sqrt[3]{u}{ }^{2}-\sqrt[3]{u} \cdot \sqrt[3]{v}+\sqrt[3]{v}^{2}\right)
$$

$\sqrt[3]{u}{ }^{2}-\sqrt[3]{u} \cdot \sqrt[3]{v}+\sqrt[3]{v}{ }^{2}=u+v$ is also rational. Substituting $(1-\sqrt[3]{u})$ for $\sqrt[3]{v}$, a quadratic equation with rational coefficients is obtained for $\sqrt[3]{u}$. Thus, $\sqrt[3]{u}=p+\sqrt{q}$, with suitable rational numbers $p, q$. From this, by cubing and rearranging, we arrive at the equation

$$
\left(3 \cdot p^{2}+q\right) \cdot \sqrt{q}=u-p^{3}-3 p \cdot q
$$

Since the number under the square root cannot be negative, in the case of positive $q$, $\left(3 p^{2}+q\right) \neq 0$. In this case, according to the equality (*), $\sqrt{q}$ is rational. Of course, this is only true if $q=0$. From this, it immediately follows that $\sqrt[3]{u}$ and $\sqrt[3]{v}$ are also rational. Therefore, the only other solutions to the original equation are $a=c \cdot e^{3}, b=c \cdot f^{3}$, where $e+f=1$ also holds. Then $c$ must be chosen so that both $a=c \cdot e^{3}$ and $b \cdot f^{3}$ are integers. These provide all other solutions.

Remarks. 1. It can be proven, albeit with much more difficulty, that the solutions to the equation $\sqrt[n]{a}+\sqrt[n]{b}=\sqrt[n]{c}$ in integers are always of the form $a=a_{1}^{n} \cdot d, b=b_{1}^{n} \cdot d, c=c_{1}^{n} \cdot d$, where $a_{1}, b_{1}, c_{1}, d_{1}$ are arbitrary integers. These can be called trivial solutions.

It can also be proven that generally, the equation

$$
c_{1}=\sqrt[n_{1}]{a_{1}}+c_{2} \cdot \sqrt[n_{2}]{a_{2}}+\cdots+c_{k} \sqrt[n_{k}]{a_{k}}=0
$$

has only "trivial" solutions in integers for $a_{i}$. The coefficients $c_{i}$ are needed here because otherwise, for example, the equality $\sqrt{2}-\sqrt{2}=0$ would be excluded. Otherwise, this is the more general case; and if it has only trivial solutions, then in every specific case, only trivial solutions exist.

It is not yet clear what we mean by a trivial solution. We must allow, for example, cases like $\sqrt[9]{8}+\sqrt[9]{-8}+\sqrt{9}+\sqrt[3]{-27}=0$. Analogously, a trivial solution can be interpreted by grouping the sum in (**). In each group, every root exponent can be "reduced" (in the case of $\sqrt[9]{8}$ and $\sqrt[9]{-8}$, the exponent is actually 3; in the case of $\sqrt{9}$ and $\sqrt[3]{-27}$, the exponent is actually 1). Now, in each group, the same $n$ exponent appears; and the numbers $a_{i}=\left(a_{i}^{\prime}\right)^{n} \cdot b$ are of the form, with integer $a_{i}$ and $b$; and for the indices appearing in the group, $\sum c_{i} \cdot a_{i}=0$ holds.

2. Among the submitters, several neglected to discuss the cases where one of the unknowns is 0, and received 4 points.