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:

1. What would be someone's thinking process when finding other formulas such as $(1)$?
2. Are there any other formulas similar to $(1)$?

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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

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.