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.
 A: 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
A: Actually, Mathematica can solve,
$$\small (x^2+axy+by^2)^3+(cx^2+dxy+ey^2)^3+(fx^2+gxy+hy^2)^3+(ix^2+jxy+ky^2)^3 =0\tag2$$
One may be guided by the principle of fait accompli (accomplished fact). Ramanujan and others already found solutions therefore $(2)$, approached the right way, must be solvable. 
What you do is expand $(2)$ and collect powers of $x,y$. The Mathematica command is Collect[P(x,y),{x,y}] to get,
$$P_1x^6+P_2x^5y+P_3x^4y^2+P_4x^3y^3+P_5x^2y^4+P_6xy^5+P_7y^6 = 0$$
where the $P_i$ are polynomials in the other variables. The hard part is then solving the system,
$$P_1 = P_2 = \dots =P_7 = 0$$
After much algebraic manipulation (which I don't have the strength to type all down), one ends up with the simple identity (which I gave to Mathworld back in 2005),

$$(ax^2-v_1xy+bwy^2)^3 + (bx^2+v_1xy+awy^2)^3 + (cx^2+v_2xy+dwy^2)^3 + (dx^2-v_2xy+cwy^2)^3 = \color{blue}{(a^3+b^3+c^3+d^3)}(x^2+wy^2)^3\tag3$$
  where,
  $$v_1= c^2-d^2\\ v_2= a^2-b^2\\ w= (a+b)(c+d)$$

Thus, if the $RHS$ is zero, or you find a single instance of $\color{blue}{a^3+b^3+c^3+d^3 = 0}$, then the $LHS$ yields a quadratic parameterization that guarantees an infinite more. So to answer your question, $(3)$ can be used to generate infinitely many Ramanujan-type formulas like $(1)$.
Example: The two smallest taxicab numbers are,
$$1^3+12^3=9^3+10^3\\ 
\color{blue}{2^3+16^3=9^3+15^3}$$ 
Using the second one and formula $(3)$, and after scaling the variable $y' \to y/12$ to reduce coefficient size, one gets,
$$(\color{blue}2 x^2 + 12 x y - 48 y^2)^3 + (\color{blue}{16} x^2 - 12 x y - 6 y^2)^3 +  (\color{blue}{-9} x^2 - 21 x y + 45 y^2)^3 + (\color{blue}{-15} x^2 + 21 x y + 27 y^2)^3  = 0$$
and you can see its "parents" in blue.
