There are lots of claims that general parametric solution of Diophantine equation $a^3+b^3+c^3=d^3$ was found but I did not find any identity or set of identities that would give all solutions.
So was the general solution really found?
$\begin{array}{cccc} \{3,4,5,6\} & \{1,6,8,9\} & \color{gray}{{6,8,10,12}} & \color{gray}{{2,12,16,18}} \\ \color{gray}{{9,12,15,18}} & \{3,10,18,19\} & \{7,14,17,20\} & \color{gray}{{12,16,20,24}} \\ \{4,17,22,25\} & \color{gray}{{3,18,24,27}} & \{18,19,21,28\} & \{11,15,27,29\} \\ \color{gray}{{15,20,25,30}} & \color{gray}{{4,24,32,36}} & \color{gray}{{18,24,30,36}} & \color{gray}{{6,20,36,38}} \\ \color{gray}{{14,28,34,40}} & \{2,17,40,41\} & \{6,32,33,41\} & \color{gray}{{21,28,35,42}} \\ \{16,23,41,44\} & \color{gray}{{5,30,40,45}} & \{3,36,37,46\} & \{27,30,37,46\} \\ \color{gray}{{24,32,40,48}} & \color{gray}{{8,34,44,50}} & \{29,34,44,53\} & \color{gray}{{6,36,48,54}} \\ \{12,19,53,54\} & \color{gray}{{27,36,45,54}} & \color{gray}{{36,38,42,56}} & \color{gray}{{9,30,54,57}} \\ \{15,42,49,58\} & \color{gray}{{22,30,54,58}} & \color{gray}{{21,42,51,60}} & \color{gray}{{30,40,50,60}} \\ \color{gray}{{7,42,56,63}} & \color{gray}{{33,44,55,66}} & \{22,51,54,67\} & \{36,38,61,69\} \\ \{7,54,57,70\} & \{14,23,70,71\} & \color{gray}{{8,48,64,72}} & \{34,39,65,72\} \\ \color{gray}{{36,48,60,72}} & \color{gray}{{12,51,66,75}} & \{38,43,66,75\} & \color{gray}{{12,40,72,76}} \\ \{31,33,72,76\} & \color{gray}{{39,52,65,78}} & \color{gray}{{28,56,68,80}} & \color{gray}{{9,54,72,81}} \\ \{25,48,74,81\} & \color{gray}{{4,34,80,82}} & \color{gray}{{12,64,66,82}} & \{19,60,69,82\} \\ \{28,53,75,84\} & \color{gray}{{42,56,70,84}} & \color{gray}{{54,57,63,84}} & \{50,61,64,85\} \\ \{20,54,79,87\} & \{26,55,78,87\} & \color{gray}{{33,45,81,87}} & \{38,48,79,87\} \\ \{21,43,84,88\} & \{25,31,86,88\} & \color{gray}{{32,46,82,88}} & \{17,40,86,89\} \\ \color{gray}{{10,60,80,90}} & \{25,38,87,90\} & \color{gray}{{45,60,75,90}} & \{58,59,69,90\} \\ \color{gray}{{6,72,74,92}} & \color{gray}{{54,60,74,92}} & \{32,54,85,93\} & \color{gray}{{15,50,90,95}} \\ \{19,53,90,96\} & \color{gray}{{48,64,80,96}} & \{45,69,79,97\} & \color{gray}{{11,66,88,99}} \\ \color{gray}{{16,68,88,100}} & \color{gray}{{35,70,85,100}} & \color{gray}{{51,68,85,102}} & \{12,31,102,103\} \\ \{33,70,92,105\} & \color{gray}{{58,68,88,106}} & \color{gray}{{12,72,96,108}} & \{13,51,104,108\} \\ \{15,82,89,108\} & \color{gray}{{24,38,106,108}} & \color{gray}{{54,72,90,108}} & \{29,75,96,110\} \\ \end{array}$
None of these (and similar ones) give them all: $$a^3 \left(a^3+b^3\right)^3=b^3 \left(2 a^3-b^3\right)^3+b^3 \left(a^3+b^3\right)^3+a^3 \left(a^3-2 b^3\right)^3\\ a^3 \left(a^3+2 b^3\right)^3=a^3 \left(a^3-b^3\right)^3+b^3 \left(a^3-b^3\right)^3+b^3 \left(2 a^3+b^3\right)^3\\ \left(4 x^2-4 x y+6 y^2\right)^3+\left(5 x^2-5 x y-3 y^2\right)^3+\left(3 x^2+5 x y-5 y^2\right)^3=\left(6 x^2-4 x y+4 y^2\right)^3$$
https://mathworld.wolfram.com/DiophantineEquation3rdPowers.html
We can read there: "The general rational solution to the 3.1.3 equation was found by Euler and Vieta". Can you provide me that solution? I do not have the books in references.
This is the general rational solution found by Euler and simplified by Binet as stated in G. H. Hardy's and E. M. Wright's book An Introduction to the Theory of Numbers.
$$\left(\lambda \left((a+3 b) \left(a^2+3 b^2\right)-1\right)\right)^3+\left(\lambda \left(1-(a-3 b) \left(a^2+3 b^2\right)\right)\right)^3=\left(\lambda \left(-\left(a^2+3 b^2\right)^2+a+3 b\right)\right)^3+\left(\lambda \left(\left(a^2+3 b^2\right)^2-a+3 b\right)\right)^3$$
It is general solution of both versions of equations (as for odd power the sign does not play any role):
$$a^3+b^3=c^3+d^3\\ a^3+b^3+(-c)^3=d^3$$