Proof that the number $\sqrt[3]{2}$ is irrational using Fermat's Last Theorem Suppose that $\sqrt[3]{2}=\frac{p}{q}$. Then $2q^3 = p^3$ i.e $q^3 + q^3 = p^3$, which is contradiction with Fermat's Last Theorem.

My question is whether this argument is a correct mathematical proof, since Fermat's Last Theorem is proven, or does it loop on itself somewhere along the proof of the Theorem? 

In other words, does the proof of Fermat's Theorem somehow rely on the fact that $\sqrt[3]{2}$ is irrational?

UPD:
As pointed out in comments, this actually is a valid argument, no matter what was used in the proof of the Fermat's Last Theorem (which from now on will be referred to as the Proof). What really interests me, is whether the Proof uses on some step the fact that $\sqrt[3]{2}$ is irrational?
In this comment BCnrd argues that this proof is "essentially circular", because converting an FLT counterexample to a Frey curve with certain congruence conditions as in the Wiles proof requires an argument equivalent to establishing irrationality of $\sqrt[3]{2}$.
does the proof of Fermat's Theorem somehow rely on the fact that $\sqrt[3]{2}$ is irrational?


Edit Added: Even if $\sqrt[3]{2}$ is irrational was contained in FLT, it would have had to be proven by some means, so as long as FLT did not assume FLT then it doesn't matter that a specific instance of FLT was contained in proof of FLT

FLT being true implies $\sqrt[3]{2}$ is irrational , ($\sqrt[3]{2}$ is irrational is a specific case of FLT).

$\sqrt[3]{2}$ is irrational is not enough to imply FLT.

as $\sqrt[3]{2}$ is irrational is a specific instance of FLT, if it was not true then it would be in contradiction of a more general theorem containing it as a specific instance of the theorem.

in other words FLT implies all specific instances of itself, where as infinitely many instances of FLT do not imply FLT.