Consider the difference between two consecutive positive cubes $f^3$ and $(f+1)^3$ is given by $3f^2 + 3f + 1$. If $f$ is a positive integer, there are no integer solutions for $\sqrt[3]{3f^2 + 3f + 1}$ unless $f=0$, as this directly follows from FLT.
I'm interested instead in the case where $f = \frac{a}{b}$ is not an integer but a rational number greater than one, and proving there are no rational cubic roots in this case. I don't think FLT applies as it is not integer, nor can I use rational root theorem. I think I have an outline of a proof that there are no rational solutions even in this case, but I am not sure if I'm approaching it correctly, so I'll outline it briefly here. First, we assume there is an irreducible rational solution $p/q$, then
$p^3 = q^3\left( \frac{3a^2}{b^2} + \frac{3a}{b} + 1 \right)$
As $p^3$ is an integer, then multiplication by integer $q^3$ must convert the RHS to an integer, so I think one can write that $q^3 = mb^2$ where $m$ is a positive whole number. It follows that
$p^3 = m(3a^2 + 3ab + b^2)$
But the bracketed term on the the RHS is an integer, so this implies that $p^3 = nm$. But this would mean that $p^3/q^3$ is not irreducible (common factor $m$), a contradiction to what we stated, suggesting there's no rational solution.
The problem is that even if this is correct, this wouldn't work in the case of $m=1$ in its current form. Is the logic of what I've done so far reasonable, and is there a way to better do this or patch the $m=1$ case?
If $\left(\frac{a}{b} + 1\right)^3 - \left(\frac{a}{b}\right)^3 = \left(\frac{c}{d}\right)^3$ for some rational numbers $a/b$ and $c/d$, you can multiply everything by $b^3d^3$ to get integers, $$ (ad + bd)^3 - (ad)^3 = (bc)^3 $$ thus but Fermat-Wiles theorem, we must have $bc = 0$ or $ad = 0$ or $(a + b)d = 0$. $b \neq 0$ and $d \neq 0$ so $a = 0$ (i.e. $f = a/b = 0$) or $a = -b$ (i.e. $f = -1$) or $c = 0$ i.e. $(f + 1)^3 = f^3$ which is impossible.