<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] Suppose $\sqrt[3]{2} = \dfrac{a}{b} \in \mathbb{Q}$ where $a, b \in \mathbb{N}_+$, $(a, b) = 1$. Now suppose $n \in \mathbb{N}_+$ satisfies$$
(\sqrt[3]{2} - 1)^n < \frac{1}{b}.$$
Note that there exist $c_0, c_1, c_2 \in \mathbb{Z}$ such that$$
(\sqrt[3]{2} - 1)^n = c_2 (\sqrt[3]{2})^2 + c_1 \sqrt[3]{2} + c_0,
$$
thus\begin{align*}
\sqrt[3]{2} &= \frac{\sqrt[3]{2} (\sqrt[3]{2} - 1)^n}{(\sqrt[3]{2} - 1)^n} = \frac{\sqrt[3]{2} (c_2 (\sqrt[3]{2})^2 + c_1 \sqrt[3]{2} + c_0)}{c_2 (\sqrt[3]{2})^2 + c_1 \sqrt[3]{2} + c_0} = \frac{c_1 (\sqrt[3]{2})^2 + c_0 \sqrt[3]{2} + 2c_2}{c_2 (\sqrt[3]{2})^2 + c_1 \sqrt[3]{2} + c_0}\\
&= \frac{c_1 \left(\dfrac{a}{b}\right)^2 + c_0 \dfrac{a}{b} + 2c_2}{c_2 \left(\dfrac{a}{b}\right)^2 + c_1 \dfrac{a}{b} + c_0} = \frac{c_1 a^2 + c_0 ab + 2c_2 b^2}{c_2 a^2 + c_1 ab + c_0 b^2},
\end{align*}
and$$
c_2 a^2 + c_1 ab + c_0 b^2 = (c_2 (\sqrt[3]{2})^2 + c_1 \sqrt[3]{2} + c_0) b^2\\
\Longrightarrow 0 < c_2 a^2 + c_1 ab + c_0 b^2 = (\sqrt[3]{2} - 1)^n b^2 < b,
$$
contradictory to the minimality of $b$. Therefore, $\sqrt[3]{2} \not\in \mathbb{Q}$.
</s>