<s>[INST] How did people get the inspiration for the sums of cubes formula? I stumbled upon this neat formula for sums of cubes with arbitrary $x,y\in\mathbb{Z}$$$(x^2+9xy-y^2)^3+(12x^2-4xy+2y^2)^3=(9x^2-7xy-y^2)^3+(10x^2+2y^2)^3\tag1$$
With $1729=1^3+12^3=9^3+10^3$ as its first instance. And I believe that this formula was used by Ramanujan to find a formula for$$a^3+b^3=c^3\pm1$$
So my question?


Questions:



*

*What would be someone's thinking process when finding other formulas such as $(1)$?

*Are there any other formulas similar to $(1)$?


I'm thinking along the lines of starting with $$(x^2+axy+by^2)^3+(cx^2+dxy+ey^2)^3=(fx^2+gxy+hy^2)^3+(ix^2+jxy+ky^2)^3$$
But even Mathematica can't solve the ensuing system that follows. So for the moment, I'm stuck.
 [/INST] The formula that Ramanujan actually recorded, as well as Euler's solution, are discussed by Ono in  https://arxiv.org/abs/1510.00735 
The original formula was
$$ \left( 6 a^2 - 4ab + 4 b^2 \right)^3 =  \left( 3 a^2 +5ab - 5 b^2 \right)^3 +  \left( 4 a^2 - 4ab + 6 b^2 \right)^3 +  \left( 5 a^2 - 5ab -3 b^2 \right)^3  $$
The formula can be made more symmetric still. The two classes of forms of discriminant $85$ are represented by
$$  x^2 + 9 xy - y^2, $$
$$  3 x^2 + 7 xy - 3 y^2.  $$ The latter is equivalent to $3 x^2 - 5 xy - 5 y^2, $ off by a single minus sign. 
The two classes of (primitive) forms of discriminant $-20$ are represented by
$$  x^2 + 5 y^2, $$
$$  2 x^2 + 2 xy + 3 y^2.  $$
That is, one may pass between Ramanujan's version and yours by using Gauss composition in order to multiply by $3,$ which passes between the principal genus and the other genus, both for discriminant $85$ and $-20.$
On page 2 they give Ramanujan's modern version of Euler's complete solution.
When
$$ \alpha^2 + \alpha \beta + \beta^2 = 3 \lambda \gamma^2, $$ 
$$  \left( \alpha + \lambda^2 \gamma  \right)^3 +  \left( \lambda \beta + \gamma \right)^3 =  \left(\lambda \alpha + \gamma \right)^3 +  \left( \beta + \lambda^2 \gamma \right)^3   $$
http://esciencecommons.blogspot.com/2015/10/mathematicians-find-magic-key-to-drive.html
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