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.

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
