Assume there exist some rationals $a, b$ such that $\sqrt[3]{a}, \sqrt[3]{b}$ are irrationals, but:
$$\sqrt[3]{a} + \sqrt[3]{b} = \frac{m}{n}$$
for some integers $m, n$
$$\implies \left(\sqrt[3]{a} + \sqrt[3]{b}\right)^3 = \frac{m^3}{n^3}$$
$$\implies a + b + 3 \cdot \sqrt[3]{ab}\left(\sqrt[3]{a} + \sqrt[3]{b}\right) = \frac{m^3}{n^3}$$
Since $a +b$, $\sqrt[3]{a} + \sqrt[3]{b}$ are rational, $\sqrt[3]{ab}$ must be rational as well.
For convenience let us say $\sqrt[3]{a} = p, \sqrt[3]{b} = q \implies pq$ is rational. This means, for all $i$, $p^iq^i$ is rational.
$$\implies (p + q)^2 = \frac{m^2}{n^2}$$
$$ = p^2 + q^2 + 2pq = \frac{m^2}{n^2}$$
Since $pq$ is rational, $2pq$ is rational and so is $p^2 + q^2$.
Assume, for some $i$ that $p^i + q^i$ and $p^{i-1} + q^{i-1}$ is rational.
$$\implies (p^i + q^i)(p + q) - pq(p^{i - 1} + q^{i - 1}) = p^{i+1} + q^{i+1}$$
is rational as well.
So for all $i$,
$$a^{\frac{i}{3}} + b^{\frac{i}{3}}$$
and
$$a^{\frac{i}{3}}b^{\frac{i}{3}}$$
are rational.
I know I'm really close to the answer, but it somehow just keeps slipping through my fingers.
As you have proved, $p^2+q^2\in\Bbb Q$ and $pq\in \Bbb Q$, so $$p-q=\frac{a-b}{p^2+pq+q^2}\in \Bbb Q.$$ Combining this with $p+q\in \Bbb Q$, we are done.