Ramanujan's sum of cubes identity is defined by the generating functions (easily calculated by Wolfram),
\begin{aligned} \sum_{n=0}^\infty a_{n} x^n &= \frac{1+53x+9x^2}{1-82x-82x^2+x^3} = 1,135,11151,\dots\\ \sum_{n=0}^\infty b_{n} x^n &= \frac{2-26x-12x^2}{1-82x-82x^2+x^3} = 2,138,11468,\dots\\ \sum_{n=0}^\infty c_{n} x^n &= \frac{2+8x-10x^2}{1-82x-82x^2+x^3} = 2,172,14258,\dots\\ \end{aligned}
then $a_n^3+b_n^3 = c_n^3 + (-1)^n$. In this 2013 post, I found the similar,
\begin{aligned} \sum_{n=0}^\infty a_n x^n &= \frac{-9(417-5602x+x^2)}{-1+184899x-184899x^2+x^3} = 3753, 693875529,\dots\\ \sum_{n=0}^\infty b_n x^n &= \frac{8(-566-11315x+x^2)}{-1+184899x-184899x^2+x^3} = 4528, 837313192,\dots\\ \sum_{n=0}^\infty c_n x^n &= \frac{-6(877+6898x+x^2)}{-1+184899x-184899x^2+x^3} = 5262, 972979926,\dots \end{aligned}
then $a_n^3+b_n^3 = c_n^3 + 1,$ and stated there should be infinitely many. Seeing this 2020 post which connected the topic to unimodular matrices made me revisit it.
I. Generating Functions
From the well-known identity,
$$(9k^4)^3+(1-9k^3) = (9k^4-3k)^3+1$$
so $(a,b,c) = (9k^4,\; 1-9k^3,\; 9k^4-3k),\;$ define,
\begin{aligned} \sum_{n=0}^\infty a_n x^n &= \frac{-a (x - 1)^2 + 144 k^4 (351 k^6 - 1) (x - 1) + 6^6 k^{10}}{x^3 - 3m x^2 + 3m x - 1}\\[4pt] \sum_{n=0}^\infty b_n x^n &= \frac{-b (x - 1)^2 - 216 k^6 (432 k^6 - 18 k^3 + 5) (x - 1) - 12^3k^6 (54 k^{6} + 1)}{x^3 - 3m x^2 + 3m x - 1}\\[4pt] \sum_{n=0}^\infty c_n x^n &= \frac{-c (x - 1)^2 - 72 k^4 (594 k^6 - 18 k^3 - 1) (x - 1) - 6^6 k^{10}}{x^3 - 3m x^2 + 3m x - 1} \end{aligned}
where $m=(432k^6 - 1)(144k^6 - 1)$ then,
$$a_n^3+b_n^3 = c_n^3 + 1\qquad$$
and the 2013 post was just the case $k=1$. If $k = -1$, then we get terms of Ramanujan's famous taxicab number $9^3+10^3 = 12^3+1 = 1729$,
\begin{aligned} \sum_{n=0}^\infty a_n x^n &= \frac{-9(417-5602x+x^2)}{-1+184899x-184899x^2+x^3} = \; 3753, 693875529,\dots\\ \sum_{n=0}^\infty b_n x^n &= \frac{-10(-323 + 9826x + x^2)}{-1+184899x-184899x^2+x^3} = -3230, -597125510,\dots\\ \sum_{n=0}^\infty c_n x^n &= \frac{-12(223 + 3664x + x^2)}{-1+184899x-184899x^2+x^3} = \; 2676, 494833692,\dots \end{aligned}
thus, $\color{blue}{3753^3 + (-3230)^3 = 2676^3 + 1}$ and so on.
II. Unimodular Matrices
To be consistent with the above, we start with $n=0$.
\begin{array}{}\begin{bmatrix} a_n \\\ b_n \\\ c_n \end{bmatrix} & = & {\begin{bmatrix} u_1v_1 & 3\times6^3\, k^7v_1 & -24\, k^3w_1 \\\ 6^3k^5u_1 & 3\times6^6\, k^{12} + 1 & - 6^3k^5u_2 \\\ 24k^3w_2 & 3\times6^3\, k^7v_2 & - u_2v_2\end{bmatrix}}^{n+1} \cdot \begin{bmatrix} 9k^4\\\ 1 - 9k^3\\\ 9k^4 - 3k\end{bmatrix}\end{array}
\begin{aligned} u_1 &= (108k^6 + 18k^3 -1)(6k^3 + 1)\\ u_2 &= (108k^6 - 18k^3 -1)(6k^3 - 1)\\ v_1 &= 216k^6 - 36k^3 -1\\ v_2 &= 216k^6 + 36k^3 -1\\ w_1 &= 81k^6(72k^6 - 36k^3 + 5)-2\\ w_2 &= 81k^6(72k^6 + 36k^3 + 5)-2\end{aligned}
which, for $n\geq0$, should yield the same values as the above so,
$$a_n^3+b_n^3 = c_n^3 + 1\qquad$$
The determinant of the $(3\times3)$ matrix, of course, is $\text{det}=1.$ For example, again let $k = -1$, so $9^3+10^3 = 12^3+1 = 1729$,
\begin{array}{}\begin{bmatrix} a_n \\\ b_n \\\ c_n \end{bmatrix} & = & {\begin{bmatrix} -111695 & -162648 & 219624 \\\ \quad 96120 & \quad 139969 & -189000\quad \\\ -79656& -115992 & 156625\end{bmatrix}}^{n+1} \cdot \begin{bmatrix} 9\\\ 10\\\ 12\end{bmatrix}\end{array}
and for $n=0$, Wolfram gives $(3753, -3230, 2676)$ thus, $\color{blue}{3753^3 + (-3230)^3 = 2676^3 + 1}$ same as above. And so on for other $n$.
III. Question
I found these relations empirically and they apparently hold true. So how do we prove them rigorously, especially the correspondence between the generating functions and the unimodular matrix?
(P.S. Note that the matrix also yields valid solutions for negative exponents $n$.)
