The Problems of Ukrainian Authors

Any group of three $(d{a}^3,d{b}^3,d{c}^3)$ if $a+b=c$ satisfies the condition of the problem.

Let's find solutions of the equation $\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}=0$. From the known equation $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$ it follows that $x+y+z=3\sqrt[3]{xyz}$, thus $(x+y+z{)}^3=27xyz$. We can assume that any two from $x$, $y$, $z$ are coprime. Really, if the prime $p$ divides $x$, $y$, then it can divide the equation $(x+y+z{)}^3=27xyz$ and therefore $z\equiv 0(\mathrm{mod}p)$.

Therefore, we can think that $x=d{x}_1$, $y=d{y}_1$, $z=d{z}_1$, where $x_1$, $y_1$, $z_1$ are coprime. Then $x_1$, $y_1$, $z_1$ are the third powers of integers $a$, $b$, $c$. On the other side, any group of three $(d{a}^3,d{b}^3,d{c}^3)$, if $a+b=c$, satisfy the condition of the problem.