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Does ZF + "For all infinite sets $A$, there is a bijection between $A^3$ and $A^2$" imply AC?

Let $\aleph(A)$ be the least ordinal that does not inject into $A$. I know $|A+\aleph(A)|=|A\times\aleph(A)|$ implies $A$ can be well ordered.

I got \begin{align}&|(A+\aleph(A))^2|\\ =&|(A+\aleph(A))^3|\\ =&|A^3+3\times A^2\times\aleph(A)+3\times A\times\aleph(A)^2+\aleph(A)^3|\\ =&|A^2+A^2\times\aleph(A)+A\times\aleph(A)+\aleph(A)|\\ \le&|4\times A^2\times\aleph(A)|\\ =&|A^2\times\aleph(A)|\\ \le&|A^2+A^2\times\aleph(A)+A\times\aleph(A)+\aleph(A)|\end{align} Since $|\aleph(A)|=|\aleph(A)^2|$, $|(A+\aleph(A))^2|=|A^2\times\aleph(A)|=|A^2\times\aleph(A)^2|=|(A\times\aleph(A))^2|$. This is almost what I need. But I don't know if this approach will work or if the statement I am trying to prove is even true.

Lucenaposition
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  • I was able to prove $|A^2\times \mu| = |A^2|$ for all $\mu<\aleph(A)$, and with some difficulty $|A^2\times \aleph(A)| = |A\times \aleph(A)|$ for all infinite $A$. I couldn't find a way to fit either of these into any of the usual proofs of Choice. My original instinct was that this won't imply Choice, but maybe my results will help someone. – Jade Vanadium Aug 05 '24 at 02:32
  • Sorry, I'm working on trying to fix the answer, but I'm traveling and my internet access is somewhat intermittent. – Asaf Karagila Aug 06 '24 at 16:03

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