elementary number theory - Proving $sqrt[3]{2}$ is irrational without using prime factorization

Prove that $\sqrt[3]{2}$ is irrational without using prime factorization.

The standard proof that $\sqrt[3]{2}$ is irrational uses prime factorization in an essential way. So I wondered if there is a proof that does not use it.

This was inspired by the fact that I know two proofs that $\sqrt{2}$ is irrational that do not use prime factorization.

The first uses
$$\sqrt{2}=\sqrt{2}\frac{\sqrt{2}-1}{\sqrt{2}-1}=\frac{2-\sqrt{2}}{\sqrt{2}-1}
$$
to show that if $\sqrt{2} = \dfrac{a}{b}$ then
$$\sqrt{2}=\frac{2-\dfrac{a}{b}}{\dfrac{a}{b}-1}=\frac{2b-a}{a-b}

$$
is a rational $\sqrt{2}$ with a smaller denominator.

The second uses
$$(x^2-2y^2)^2=(x^2+2y^2)^2-2(2xy)^2
$$
and $3^2-2\cdot 2^2 = 1$ to show that $x^2-2y^2=1$ has arbitrarily large solutions and this contradicts $\sqrt{2}$ being rational.

I have not been able to extend either of these proofs to $\sqrt[3]{2}$. Results that I do not consider "legal" in solving this problem include Fermat's Last Theorem (which definitely uses unique factorization) and the rational root theorem (which uses unique factorization
in its proof).