I came across this identity $$\left(x^2+7xy-9y^2\right)^3+\left(2x^2-5xy+12y^2\right)^3=\left(2x^2+10y^2\right)^3+\left(x^2-9xy-y^2\right)^3$$
Is there a deeper reason why a formula such as this should exist?
Is there some background where the search for it could be motivated, and that it would come naturally?
(Updated with second cubic case.) Sorry for the late reply, I was inactive for about 4 years. Anyway, most of Ramanujan's results are special cases of more general and perhaps deeper phenomena.
I. Third powers
A. General case
For your question, a more general algebraic identity is,
$$A^3+B^3+C^3+D^3 = (a^3+b^3+c^3+d^3)(x^2+wy^2)^3$$
and $(A,B,C,D)$ are quadratic forms,
\begin{align} A &= ax^2-v_1xy+bwy^2\\ B &= bx^2+v_1xy+awy^2\\ C &= cx^2+v_2xy+dwy^2\\ D &= dx^2-v_2xy+cwy^2 \end{align}
where {$v_1,\, v_2,\, w$} = {$c^2-d^2,\; a^2-b^2,\; (a+b)(c+d)$}. I gave this identity to Mathworld about 20 years ago.
Thus, given a single example of $a^3+b^3+c^3+d^3 =N$ where $N=0$, a cube, or any number of cubes, then one can generate an infinite more via the LHS. For example, try it with the well-known $3^3+4^3+5^3+(-6)^3 = 0$, permute these variables $(a,b,c,d)$ in any order you wish, and you'll have your own Ramanujan-type identity of sums of cubes.
B. Special case (Update, Sept. 13, 2024)
Given a solution to $a^3+b^3+c^3+d^3 = e^3$, with $a+b = c+d$, then similarly,
$$A^3+B^3+C^3+D^3 = (a^3+b^3+c^3+d^3)(x^2+y^2)^3$$
and quadratic forms,
\begin{align} A &= ax^2+u_1xy+by^2\\ B &= bx^2-u_1xy+ay^2\\ C &= cx^2-u_2xy+dy^2\\ D &= dx^2+u_2xy+cy^2 \end{align}
where {$u_1,\, u_2 = c-d,\, a-b$}. Notice that $A+B = C+D$. For example, given the nice,
$$11^3+14^3+12^3+13^3 = 20^3$$
with $11+14=12+13$, and many others. Then,
$$A^3+B^3+C^3+D^3 = 20^3(x^2+y^2)^3$$
where $(A,B,C,D)$ are the polynomials above.
II. Fourth powers
Ramanujan also gave two quadratic parametrizations for,
$$a^4+b^4+c^4+d^4+e^4 = f^4$$
where $f$ is always divisible by $5$, one of which is,
$$(2x^2+12xy-6y^2)^4+(2x^2-12xy-6y^2)^4+(4x^2-12y^2)^4+(4x^2+12y^2)^4+(3x^2+9y^2)^4 = 5^4(x^2+3y^2)^4$$
Notice the similar relation $a+b=c$ to the special cubic case. For $(x,y)=(1,0)$ this yields the nice,
$$2^4+2^4+4^4+4^4+3^4 = 5^4$$
We can find infinitely more polynomials. The general identity (which I also gave to Mathworld) is let $a+b=c$, then,
$$A^4+B^4+C^4 = (a^4+b^4+c^4)(x^2+3y^2)^4$$
and $(A,B,C)$ are also quadratic forms,
\begin{align} A &= ax^2+2(b+c)xy-3ay^2\\ B &= bx^2-2(a+c)xy-3by^2\\ C &= cx^2-2(a-b)xy-3cy^2\\ \end{align}
Thus a single $a^4+b^4+c^4+d^4+e^4 =f^4$ where $a+b = c$ yields a polynomial parameterization. For example,
$$50^4+50^4+100^4+4^4+15^4 = 103^4$$
therefore,
$$A^4+B^4+C^4 = (-4^4-15^4+103^4)(x^2+3y^2)^4$$
$$A^4+B^4+C^4+4^4(x^2+3y^2)^4+15^4(x^2+3y^2)^4 = 103^4(x^2+3y^2)^4$$
where $(A,B,C)$ are the polynomials above, so $f$ is always divisible by $103$, and not found by Ramanujan.
P.S. There's actually another cubic identity, but I'm too sleepy. I'll add it soon. (Added a year later, Sept 13, 2024.) More details can be found at "Generalizing Ramanujan's sum of cubes identity?"
This is one of many Number Theoretic identities that mathematician Ramanujan discovered. It's hard to say exactly what the "deeper reason" such a formula exists, but it was most likely discovered in search of generalized families of Taxicab numbers. The formula shows a method to write a number as in two distinct ways as a sum of two separate cube numbers. For example: \begin{align*} 1729 &= 1^3 + 12^3 \\ &= 9^3 + 10^3 \\ 87539319 &= 167^3 + 436^3 \\ &= 228^3 + 423^3 \\ &= 255^3 + 414^3. \end{align*} (See if you can find which $x,\: y$, and $z$ you need to reach these solutions!)
As a few commentors have pointed out, Ramanujan claimed to have received many formulas like these from the goddess Namagiri, the Hindu Goddess of creativity. It's an interesting bit of mathematical history that I think you would be delighted to learn. Ramanujan was truly an incredible mind.
