sum of cubes of two rationals How to find two rational numbers $x,y$ such that $$x^3+y^3=6$$ I know that $x=17/21,y=37/21$ is a solution but I am interested in a method how is achieved and does exists other solutions 
 A: 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.
A: 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 
A: I have used Microsoft Solver Foundation to find a (different) solution:
SolverContext context = SolverContext.GetContext();

Decision a = new Decision(Domain.IntegerNonnegative, "A");
Decision b = new Decision(Domain.IntegerNonnegative, "B");
Decision c = new Decision(Domain.IntegerNonnegative, "C");
Decision d = new Decision(Domain.IntegerNonnegative, "D");

Model model = context.CreateModel();
model.AddDecisions(a, b, c, d);

Term a3 = a * a * a;
Term b3 = b * b * b;
Term c3 = c * c * c;
Term d3 = d * d * d;

Term res = a3 * d3 + c3 * b3 - 6 * b3 * d3;

model.AddConstraint("eq", res == 0);
model.AddConstraint("a1", a < 1000000);
model.AddConstraint("b1", b < 1000000);
model.AddConstraint("c1", c < 1000000);
model.AddConstraint("d1", d < 1000000);
model.AddConstraint("a2", a >= 1);
model.AddConstraint("b2", b >= 1);
model.AddConstraint("c2", c >= 1);
model.AddConstraint("d2", d >= 1);
// model.AddConstraint("a3", a > c);  //  symmetry breaking

model.AddConstraint("b3", b != 21);   //  want something different!

Solution solution = context.Solve();

Console.WriteLine("a={0} b={1} c={2} d={3}", a, b, c, d);

The solver re-discovers your solution in a couple of seconds but is unable to find a different one with numbers below 1000000.
A: Note the danger that most solutions found by the Julia program will reduce to 17/21 and 37/21 for a and b or the reverse. However If this program uses BigInt for the integers it will find another unique solution in a week to two. You also can test and skip any a/c that reduces to 17/21 with the test
    if a//c == 17/21; continue; end 

before the print statement and declare a,b,c BigInt and increase the range of c by another decade.
