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

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. 
