<s>[INST] 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).
 [/INST] 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.
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