$1729$, Fermat's Last Theorem, and Ramanujan's sums of cubes formula In the following page in one of Ramanujan's Lost Notebooks, Ramanujan found a formula for sums of cubes such as the famed $1729$.

Which can be found in the bottom right hand corner. But also in that page, is a formula:

If$$\sum_{n\geq0}a_nx^n=\frac {1+53x+9x^2}{1-82x-82x^2+x^3}\\\sum_{n\geq0}b_nx^n=\frac {2-26x-12x^2}{1-82x-82x^2+x^3}\\\sum_{n\geq0}c_nx^n=\frac {2+8x-10x^2}{1-82x-82x^2+x^3}$$
  Then$$a_n^3+b_n^3=c_n^3+(-1)^n$$
My Question: How would you go about proving this?

I did notice one thing: It seemed like Ramanujan was using generating functions. Where given a sequence, you pretend that the numbers are coefficients of a polynomial, and you can then collapse that down into a single expression.
For example: The counting numbers $1,2,3,4,5,\ldots$ can be represented as$$1+2x+3x^2+4x^3+\ldots=\frac {1}{(1-x)^2}\tag{1}$$
But if you use generating functions, then you can't substitute that $x$ with anything, which makes them both powerful,  but dangerous. So what would be the point of using generating functions?

Michael Hirschhorn has given two proofs of Ramanujan's cube equations:
Michael D. Hirschhorn, An amazing identity of Ramanujan, Mathematics Magazine 68 (1995) 199–201.
Michael D. Hirschhorn, A proof in the spirit of Zeilberger of an amazing identity of Ramanujan, Mathematics Magazine 69 (1996) 267–269.
For a short summary see also here. In fact, the existence of this result by Ramanujan depends very much on special circumstances of the solution
$$
(A^2 +7AB−9B^2)^3 +(2A^2 −4AB+12B^2)^3 = (2A^2 +10B^2)^3 +(A^2 −9AB−B^2)^3,
$$
for $4$ cubes.
Tito Piezas has found several other Ramanujan-like families of such identities with four cubes, see this MSE-question.
