A Ramanujan-type identity: $11\sum_{n=1}^{\infty}\frac{n^3}{e^{2n\pi}-1}-16\sum_{n=1}^{\infty}\frac{n^3}{e^{4n\pi}-1}=\frac{1}{48}$ Out of curiosity, why it is these sums yield a rational answer?
$$11\sum_{n=1}^{\infty}\frac{n^3}{e^{2n\pi}-1}-16\sum_{n=1}^{\infty}\frac{n^3}{e^{4n\pi}-1}=\frac{1}{48}$$
I found this identity during the observing ramanujan identiy via a wolfram sum calculator.
Or it is also related to the type of $ e^{\pi{\sqrt43}}=884736743.999$ it is almost an integer?
Also,if possible can anyone verify it is correctness?

I provide an approach which involves many identities from the classical theory of elliptic and theta functions. Also I present them using Ramanujan's notation. Although the theory of elliptic and theta functions needs complex analysis to understand all the details, the formulas which I give here can be established within limits of real analysis only (and the proofs are available in my blog posts, let me know if you need them).
In what follows $k$ is a real number with $0 < k < 1$ and $k' = \sqrt{1 - k^{2}}$. Further $$K = K(k) = \int_{0}^{\pi/2}\frac{dx}{\sqrt{1 - k^{2}\sin^{2}x}}, K' = K(k'), E = E(k) = \int_{0}^{\pi/2}\sqrt{1 - k^{2}\sin^{2}x}\,dx$$ and $$q = \exp\left(-\pi\cdot\frac{K'}{K}\right)$$

Let us use Ramanujan Function $$Q(q) = 1 + 240\sum_{n = 1}^{\infty}\frac{n^{3}q^{n}}{1 - q^{n}}\tag{1}$$ Then we know that $$Q(q^{2}) = \left(\frac{2K}{\pi}\right)^{4}(1 - k^{2} + k^{4})\tag{2}$$ where $q = e^{-\pi K'/K}$ and we can change $q$ to $q^{2}$ if $k$ is replaced by $(1 - k')/(1 + k')$ and $K$ is replaced by $K(1 + k')/2$ Thus we have $$Q(q^{4}) = \left(\frac{K(1 + k')}{\pi}\right)^{4}\left(1 - \frac{(1 - k')^{2}}{(1 + k')^{2}} + \frac{(1 - k')^{4}}{(1 + k')^{4}}\right)$$ or $$Q(q^{4}) = \left(\frac{2K}{\pi}\right)^{4}\left(1 - k^{2} + \frac{k^{4}}{16}\right)\tag{3}$$ Lets put $q = e^{-\pi}$ to get $K' = K$ so that $k = 1/\sqrt{2}$. Then $$\sum_{n = 1}^{\infty}\frac{n^{3}}{e^{2n\pi} - 1} = \frac{Q(q^{2}) - 1}{240}, \sum_{n = 1}^{\infty}\frac{n^{3}}{e^{4n\pi} - 1} = \frac{Q(q^{4}) - 1}{240}$$ so the desired sum is $$\frac{1}{240}\{5 + 11Q(q^{2}) - 16Q(q^{4})\} = \frac{1}{48} + \frac{(2K/\pi)^{4}}{240}\{11\cdot(3/4) - 16\cdot(33/64)\} = \frac{1}{48}$$

About the values of such sums being rational I don't see them as surprising. Clearly for $q = e^{-\pi}$ we have $$Q(q^{2}) = \frac{3}{4}\left(\frac{2K}{\pi}\right)^{4}$$ And we can connect it with many other formulas like $$P(q) = 1 - 24\sum_{n = 1}^{\infty}\frac{nq^{n}}{1 - q^{n}} = \left(\frac{2K}{\pi}\right)^{2}\left(\frac{6E}{K} + k^{2} - 5\right)$$ and $$P(q^{2}) = \left(\frac{2K}{\pi}\right)^{2}\left(\frac{3E}{K} + k^{2} - 2\right)$$ so that $$2P(q^{2}) - P(q) = \left(\frac{2K}{\pi}\right)^{2}\left(1 + k^{2}\right) = \frac{3}{2}\left(\frac{2K}{\pi}\right)^{2}$$ and hence finally $$3Q(q^{2}) - \{2P(q^{2}) - P(q)\}^{2} = 0$$ And this will give a polynomial equation with rational coefficients connecting $P(q), P(q^{2}), Q(q^{2})$. Many such relations can be derived via simple manipulation of the formulas for $P, Q, R$ in terms of $k, K, E$. Just to add one last example of such sums with rational value, it can be established using some of Ramanujan's results (or some other techniques which I am not aware of) that $$\sum_{n = 1}^{\infty}\frac{n^{s}}{e^{n\pi} - 1}$$ is rational for all positive integers $s$ of the form $s = 4k + 1$ (the sum has a closed form for odd values of $s$ but not for even values of $s$).
In his paper "Modular Equations and Approximations to $\pi$" Ramanujan does the ultimate form of cancellation to obtain various series for $1/\pi$ some of which consists purely of rational terms and the most spectacular one consists of just one irrational $\sqrt{2}$ namely $$\boxed{\displaystyle \frac{1}{2\pi\sqrt{2}} = \frac{1103}{99^{2}} + \frac{27493}{99^{6}}\frac{1}{2}\frac{1\cdot 3}{4^{2}} + \frac{53883}{99^{10}}\frac{1\cdot 3}{2\cdot 4}\frac{1\cdot 3\cdot 5\cdot 7}{4^{2}\cdot 8^{2}} + \cdots}$$ See the series on my blog starting with this post for more details.
