355/113 and small odd cubes

An important approximation to $\pi$ is given by the convergent $\frac{355}{113}$.

The numerator and the denominator of this fraction are at the same distance of small consecutive odd cubes.

$$\frac{355}{113} = \frac{7^3+12}{5^3-12}$$

Is this a consequence of some general formula, such as a series or continued fraction?

Attempts, drafts

$$\frac{113}{355} = \frac{1}{10}\left(2+\frac{5^3}{3+\frac{7^3}{4}}\right)-\frac{8}{355} = \frac{1}{5}\left(3-\frac{5^3}{3+\frac{7^3}{4}}\right)$$

Similarly, for $\frac{22}{7}$

$$\frac{7}{22} =\frac{1}{2}\left(\frac{3^3}{1+\frac{4^3}{2}}\right) -\frac{1}{11}$$

For the first convergent $3$, $$\frac{3^3-6}{1^3+6}=\frac{21}{7}=3$$

The tentative expression $$a(n)=\frac{(4n+3)^3+6(3n-1)}{(4n+1)^3-6(3n-1)}$$

has $a(0)=3$ and $a(1)=\frac{355}{113}$ but $\lim_{n \to \infty} a(n) = 1 \neq \pi$

A similar fraction for the approximation $\frac{223}{71}$ (Archimedes' lower bound) is $$\frac{98+5^3}{98-3^3}=\frac{2·7^2+5^3}{2·7^2-3^3}=\frac{223}{71}=3+\frac{1}{7+\frac{1}{10}}$$

For a coprime pair of integers $1 \le p,q$, consider the following diophantine equation: $$\frac{x^3+z}{y^3-z}= \frac{p}{q}$$ with $x,y,z$ integer unknowns. This is equivalent to $$z= \frac{y^3p-x^3q}{p+q}$$ Thus a solution exists if and only if there exist $x,y$ such that $$y^3p-x^3q \equiv 0 \pmod{(p+q)}$$ Now, note that $-q \equiv p$, and that it has an inverse $\mod (p+q)$. This means that we need $$x^3+y^3 \equiv 0 \pmod{(p+q)}$$

Now, if we pick $p=355$ and $q=113$, we get $$p+q=468 = 7^3+5^3$$ In other words, the fraction $355/113$ has such a special property not because it is a convergent of $\pi$, but simply because the sum of numerator and denominator is a sum of two small cubes. We can do the same job with the fraction $$\frac{203}{265} = \frac{5^3+78}{7^3-78}$$ nothing special with $\pi$. As for $22/7$ we can find $$\frac{22}{7} = \frac{17^3+5^3}{12^3-5^3}$$ which is even more surprising. In my opinion, having three variables gives you a lot of freedom, so that yo can find such nice forms for any rational number.

EDIT: Just for fun, I looked up convergents of a completely different constant, namely $\gamma$: Euler-Mascheroni constant. One of its convergents is $$\frac{71}{123}= \frac{3^3+470}{11^3-470}$$ Again, there is nothing special about this rational, simply you can find such a form for any number.