Any group of three $(da^3, db^3, dc^3)$ 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 $z \equiv 0 \pmod{p}$.

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 $(da^3, db^3, dc^3)$, if $a + b = c$, satisfy the condition of the problem.