<s>[INST] Ramanujan, sum of two cubes - how was it discovered? I'm looking for motivation for, and hopefully a derivation of, Ramanujan's sum of  cubes formula 
$$\left(x^2+7xy-9y^2\right)^3+\left(2x^2-4xy+12y^2\right)^3=\left(2x^2+10y^2\right)^3+\left(x^2-9xy-y^2\right)^3$$
I can't see where one could start to justify this without actually expanding everything out. 
Additionally, does it provide a parametrisation of the surface 
$$X^3+Y^3=Z^3+1$$ in $\mathbb{Q}^3$
with 
$$X=\frac{x^2+7xy-9y^2}{2x^2+10y^2}$$
$$Y=\frac{2x^2-4xy+12y^2}{2x^2+10y^2}$$
$$Z=\frac{x^2-9xy-y^2}{2x^2+10y^2}$$
 [/INST] The most general homogeneous quadratic function of two variables is $ax^2+bxy+cy^2$. Cubing this gives $$a^3x^6+3a^2bx^5y+(3a^2c+3ab^2)x^4y^2+(b^3+6abc)x^3y^3+(3ac^2+3b^2c)x^2y^4+3bc^2xy^5+c^3y^6.$$(The fact that this sextic is homogeneous makes the above easy to deduce from combinatorics, without manually expanding out brackets. There's even an $x\leftrightarrow y$ symmetry to almost halve the work.) The sum of two such cubes equals another viz.$$(ax^2+bxy+cy^2)^3+(dx^2+exy+fy^2)^3=(gx^2+hxy+iy^2)^3+(jx^2+kxy+ly^2)^3$$iff coefficients match. The first requirement this gives is$$a^3+d^3=g^3+j^3.$$Ramanujan was famous for noting the least positive integer expressible as the sum of two cubes in two different ways is $1729=1^3+12^3=9^3+10^3$, but in this example we work with much smaller examples, so we just swap variables. The "simplest" option is $a=1,\,d=2$ (so that $d\ne a$), whence $g=d=2,\,j=a=1$. For the $y^6$ coefficients, Ramanujan actually does work with the above result about $1729$, rearranging the equality of two sums of cubes as$$(-9)^3+12^3=10^3+(-1)^3$$to reduce the size of the total to $999$. So we try $c=-9,\,f=12,\,i=10,\,l=-1$. We now only need to find $b,\,e,\,h,\,k$, which in the case at hand turn out to be $7,\,-4,\,0,\,-9$. History probably doesn't record whether Ramaujan found these through trial and error or by working with the other coefficient-matching conditions, but I'm sure he did the latter, and you can try it as an exercise. (Remember to plug in the $8$ coefficients we've already fixed to simplify it to a problem in the other $4$.)
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