sum of cubes of two rationals

Solutions $z$ of the diophantine equation $x^3 + y^3 = 6z^3$ are tabulated at the Online Encyclopedia of Integer Sequences. Only $4$ are given (though infinitely many exist): $21$, $960540$, $16418498901144294337512360$, and $436066841882071117095002459324085167366543342937477344818646196279385$ $305441506861017701946929489111120$.

See also this mathforum post, and the article, The £$450$ question, by J. H. E. Cohn, Mathematics Magazine 73, No. 3 (Jun., 2000) 220-226.

EDIT: Indeed, Cohn gives a solution not in the OEIS, and smaller than that last solution: $$z=1097408669115641639274297227729214734500292503382977739220$$ It's a very nice paper.

Using the maple syntax from this site,
I have here $6$ $z$ such that: $$ x^3 + y^3 =6z^3 $$ I have excluded the other $z$'s for the $7^{\text{th}}$ is nearly $30,000$ digits long.

link

This is an old question, but anyway. Given an initial solution $x,y,z$, to,

$$ax^3+by^3 = cz^3$$

then a new one can be derived as,

$$a(-bxy^3-cxz^3)^3 + b(ax^3y+cyz^3)^3 = c(-ax^3z+by^3z)^3\tag{0}$$

For example, given the OP's,

$$x^3+y^3 = 6z^3$$

starting with initial,

$$x,y,z = 17, 37, 21\tag{1}$$

using $(0)$, we find a second,

$$x,y,z = -1805723,\, 2237723,\, 960540\tag{2}$$

which is the point given by Myerson and Yazdanpour. Using $(2)$, we can find a third and so on, ad infinitum.

P.S. 1. Presumably, a positive $x,y,z$ will appear after every few iterations. 2. For some reason, the solution given by Kohn is skipped by this process.