A Paradox of Infinite Long Division // Un paradosso della divisione infinita

 Everything started from the aspects provided in the Post "An easy...".(I still haven't found a simple explanation.)

         I will extract from there the equation 

$26^3-24^3-12^3+1^3=(2^3+1)\cdot 225$

which expresses that  $2025=9\cdot 225$. I'm thinking of seeing what a writing by 225 looks like, which maybe I haven't taken into account yet. I force a division for this,  $26^3-24^3-12^3+1\;:\;(2^3+1)$ , imitating the long division of two polynomials.

\begin{array}{ccccccc}\;\;\;26^3&-24^3&\;&-12^3&+1&\vdash&2^3+1\\-26^3&\;&-13^3&&&&13^3-12^3-\left(\frac{13}{2}\right)^3+\left(\frac{13}{4}\right )^3-\left (\frac{13}{8}\right)^3+\left (\frac{13}{16}\right)^3-\left (\frac{13}{32}\right )^3+\cdots\\\hline \setminus&-24^3&-13^3&-12^3&+1\\\;&24^3&\;&12^3\\\hline \;&\setminus&-13^3&\setminus&+1\\\;&\;&13^3&+\left(\frac{13}{2}\right)^3\\\hline \;&\;&\setminus&\left(\frac{13}{2}\right)^3&\;&+1\\\;&\;&\;&-\left(\frac{13}{2}\right)^3&-\left(\frac{13}{4}\right)^3\\\hline \;&\;&\;&\setminus&-\left(\frac{13}{4}\right)^3&+1\\\;&\;&\;&\;&\left(\frac{13}{4}\right)^3+\left (\frac{13}{8}\right)^3\\\hline \;&\;&\;&\;&\left (\frac{13}{8}\right)^3&+1\\\;&\;&\;&\;&-\left(\frac{13}{8}\right)^3-\left(\frac{13}{16}\right)^3\\\hline \;&\;&\;&\;&-\left(\frac{13}{16}\right)^3&+1\\\;&\;&\;&\;&\left(\frac{13}{16}\right)^3+\left(\frac{13}{32}\right)^3\\\hline \;&\;&\;&\;&\left(\frac{13}{32}\right)^3&\color{Red}{+1}\\\:&\;&\;&\;&\dots&\dots\end{array}

The division can continue indefinitely. The quotient of this division is

$13^3-12^3-\left ( \frac{13}{2}\right )^3+\left ( \frac{13}{4}\right )^3-\left ( \frac{13}{8}\right )^3+\left (\frac{13}{16}\right )^3-\left ( \frac{13}{32}\right )^3+ \cdots=$

$=-12^3+13^3 \cdot \left [1-\left (\frac{1}{2}\right )^3+\left (\frac{1}{2}\right )^6-\left (\frac{1}{2} \right )^9+\left (\frac{1}{2} \right )^{12}-\left (\frac{1}{2} \right)^{15}+\cdots \right ]=$

$=13^3\cdot \frac{1}{1+\frac{1}{8}}-12^3=\frac{8}{9}\cdot 13^3-12^3=\frac{26^3}{9}-12^3.$

We would have expected the result to be 225, so 

$\color{Red}{\frac{26^3}{9}-12^3=225}$

But this is NOT true, and in this I considered, mistakenly, that it was a paradox. We must take into account that for each partial remainder of the division there remains a  $\color{Red}{+1}$. The correct result of the division is

$\color{Blue}{\frac{26^3}{9}-12^3+\frac{1}{9}}$

          I asked ChatGPT what other possibilities there were to write the number 225 as a sum of cubes and he gave me the following reasonable results, but with which I made no progress at all :

$6^3+2^3+1^3$

$7^3-4^3-3^3-3^3$

$7^3-6^3+5^3-3^3$

$7^3-5^3+2^3-1^3$

$8^3-7^3+4^3-2^3$

$9^3-8^3+2^3$

$10^3-7^3-6^3-6^3$