If $n$ is a positive integer, $a$, $b$, and $\sqrt[n]{a}-\sqrt[n]{b}$ are rational, then each of $\sqrt[n]{a}$ and $\sqrt[n]{b}$ must be rational. Assume that $a$ and $b$ are rational numbers with $a\neq b$ such that $\sqrt{a}-\sqrt{b}$ is rational. The each of $\sqrt{a}$ and $\sqrt{b}$ must be rational. Because if $\sqrt{a}-\sqrt{b}$ is rational so it is its reciprocal which is $$\frac{\sqrt{a}+\sqrt{b}}{a-b}.$$ This implies that $\sqrt{a}+\sqrt{b}$ must be rational as must $2{\sqrt a}$ and $2{\sqrt b}$, and hence each of $\sqrt{a}$ and $\sqrt{b}$ must be rational.
I am trying to prove the following assertion:
Assume that $a$ and $b$ are rational numbers such that $\sqrt[3]{a}-\sqrt[3]{b}$ is rational. The each of $\sqrt[3]{a}$ and $\sqrt[3]{b}$ must be rational.
More generally, if $a$ and $b$ are rational numbers such that $\sqrt[n]{a}-\sqrt[n]{b}$ is rational for any positive integer $n$, then each of $\sqrt[n]{a}$ and $\sqrt[n]{b}$ must be rational.
For the case $n=3$, I was able to conclude that since $$a-b =(\sqrt[3]{a}-\sqrt[3]{b})(\sqrt[3]{a^2}+\sqrt[3]{a}\sqrt[3]{b}+\sqrt[3]{b^2})$$ is rational, it must be the case that $$\sqrt[3]{a^2}+\sqrt[3]{a}\sqrt[3]{b}+\sqrt[3]{b^2}$$ must be rational. Not sure where to go from here!

Easiest is to say the following. Assume $x^k-a$ and $x^l-b$ are irreducible over $\mathbb{Q}$ then $\sqrt[k]{a}-\sqrt[l]{b}$ is irrational, for otherwise if $k\leq l$,
$$a=(\sqrt[l]{b}+r)^k$$ is an $k$th degree equation for $\sqrt[l]{b}$, or $k-1$ degree if they are equal.
Note that if $$x^n-a=f(x)g(x)$$ is reducible with the degree of $f$ being $k$ then the roots of $f$ are of the form $\epsilon \sqrt[n]{a}$ and their product, the constant coefficient of the form $\eta(\sqrt[n]{a})^k\in \mathbb{Q}$ which implies that $(\sqrt[n]{a})^k\in \mathbb{Q}$ and so $a=q^{n/k}$. This is one case of a much more general theorem proved by other means.
