Is the difference of two irrationals which are each contained under a single square root irrational? Is $ x^\frac{1}{3} - y^\frac{1}{3}$ irrational, given that both $x$ and $y$ are not perfect cubes, are distinct and are integers (i.e. the two cube roots are yield irrational answers)?
I understand that the sum/difference of two irrationals can be rational (see this thread: Is the sum and difference of two irrationals always irrational?). However, if my irrationals are contained under one root (so for example $3^\frac{1}{3}$ and not $2^\frac{1}{2} + 1$), can one generalise to show that $ x^\frac{1}{p} - y^\frac{1}{q} $ is irrational, where of course $x$ and $y$ are not powers of $p$ and $q$ respectively?

Denote the cube roots by $X,Y$, so that $X^3=x$ and $Y^3=y$ with $x,y\in \mathbb Z$.
Suppose, with slightly greater generality, that we have $X-Y-R=0$ where $X^3,Y^3,R\in \mathbb Q$.
We first want to argue that $XY\in \mathbb Q$.  To do so, observe the identity:
$$X^3-Y^3-R^3-3XYR=(X-Y-R)(X^2+Y^2+R^2+XY+XR-YR)$$
This immediately implies that $$3XYR=X^3-Y^3-R^3$$  which, as desired, implies that $XY\in \mathbb Q$.
But then we have two real numbers, $X,Y$ such that both $X-Y$ and $XY$ are rational.  It follows that $X,-Y$ both satisfy a quadratic equation with rational coefficients, namely
\begin{align}
X^2 - (X-Y)X -XY &= 0,& Y^2 +(X-Y)Y -XY&=0.
\end{align}
But this is not possible, as $Z^3-x=0$ is the minimal polynomial for the cube root of $x$ over $\mathbb Q$.
