Integer solutions to the diophantine equation $a^3+b^3+c^3=6$

Question

I once thought that this equation $x^3+y^3+z^3=6$ has only a few smaller integer solutions. Until somebody told me this $6=192722201207819^3+162765491944499^3+(-225522344776678)^3$ existed. I don't know how it came about.

Question: How to find more integer solutions to the equation $a^3+b^3+c^3={\color{red}{6}}$?

a few smaller integer solutions: \begin{align*} (-637)^3+(-205)^3+644^3&=6\\ (-235)^3+(-55)^3+236^3&=6\\ (-58)^3+(-43)^3+65^3&=6\\ (-1)^3+(-1)^3+2^3&=6\\ \end{align*}

Something happened recently

I've also heard that mathematicians have recently solved this problem:

The least integer solutions to the equation $a^3+b^3+ c^3=33$

Triple cubic Diophantine equations

42 is the new 33 - Numberphile

@YuiToCheng Sorry, I didn't make it clear. It has been added now. — D.Matthew, Mar 15 '19 at 13:52
First of all, the fact that a similar question was solved for $33$ seems to be irrelevant. (Unless the opposite is proved) Secondly, What did you try? — Arman Malekzadeh, Mar 15 '19 at 14:04
The sums of three cubes problem is open. It is unclear whether your interest is limited to ways of expressing six (as the body of the Question suggests) or about the broad family of problems (as your links and closing remarks hint at). — hardmath, Mar 15 '19 at 15:29

score 3 · Answer 1 · answered Mar 15 '19 at 13:39

Nothing more than brute force computation I'm afraid.

The method used to crack $33$ was to rearrange to $$(a^2-ab+b^2)=\frac{33-c^3}{a+b}$$ The computer can then assign a random $c$ and test all values of $a,b$ such that $(a+b)|(33-c^3)$ which requires much less testing, but still very large amounts.

Integer solutions to the diophantine equation $a^3+b^3+c^3=6$

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