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).

You can do it using the concept of odds and evens which is sort of a poor mans version of unique prime factorization but surely acceptable.  After all the classic proof of the irrationality of $\sqrt 2$ used even and odds without assuming unique prime factorization.
First note:  If $b \in \mathbb Z$ is $b = 2m$ is even, then $b^3 = 8m^3$ is even and if $b = 2m + 1$ is odd then $b^3 = 8m^3 + 12m^2 + 6m + 1$ is odd.
So if $\frac ab; b\ne 0$ and $a$ and $b$ in "lowest terms" (have no factors in common, in particular are not both even), and if $(\frac ab)^3 = 2$ then....
$2a^3 = b^3$ and so $b^3$ is even and so $b$ is even and so $b = 2m$ and so $2a^3 = 8m^3$ and so $a^3 = 4m^3$ and so $a^3$ is even and so $a$ is even and so $a$ and $b$ are both even but we said that wasn't the case so that's impossible and nyah nyah nyah nyah nyah.
