Cute double sum

Let $\displaystyle \xi=\sum_{n=1}^{\infty}\left(\sum_{k=1}^n \left(\frac {1}{(25k^2+25k+1)(n+1-k)^3}\right)\right)$

Now writing the the total sum in different rows for each value of $n$ or interchanging the sums we get $\displaystyle \xi =\sum_{n=1}^{\infty} \frac {1}{n^3}\sum_{k=1}^{\infty}\frac {1}{25k^2+25k+1}=\zeta(3)\sum_{k=1}^{\infty} \frac {1}{25k^2+25k+1}$

$\xi =\frac {\zeta(3)}{3}\sum_{k=1}^{\infty}\left(\frac {1}{5k+1}-\frac {1}{5k+4}\right)$

Let $\displaystyle \xi_1=\sum_{k=1}^{\infty}\left(\frac {1}{5k+1}-\frac {1}{5k+4}\right)=\sum_{k=1}^{\infty}\int_0^1 (1-x^3)x^{5k}dx$

Interchanging the integral and the sum which is justified by Fubini's theorem and then using Geometric sum $\displaystyle \xi_2=\int_0^1 (1-x^3)\frac {x^5}{1-x^5}dx$

Substituting $x^5=t$ in $\xi_2$ we get

$\displaystyle \xi_1=\frac 15\int_0^1 \frac {t^{\frac 15}-t^{\frac 45} +1-1}{1-t}dt$

Using the relation of Digamma function that $\displaystyle \psi(z+1)+\gamma=\int_0^1 \frac {1-x^z}{1-x}dx$

we get $\displaystyle \xi_1=\frac 15\left(\psi\left(\frac 95\right)-\psi\left(\frac 65\right)\right)$ Using the functional rule of Digamma function that $\displaystyle \psi(z+1)=\psi(z)+\frac 1z$ we get $\displaystyle \xi_1=\frac 15\left(\psi\left(\frac 45\right)-\psi\left(\frac 15\right)-\frac {15}{4}\right)$ Using the Reflection formula of Digamma function that $\displaystyle \psi(1-z)-\psi(z)=\pi\cot(\pi z)$ we get that $\displaystyle \xi_1=\frac 15\left(\pi\cot\left(\frac {\pi}{5}\right) -\frac {15}{4}\right)$

Using this alongwith result of $\xi$ we get $\displaystyle \xi=\frac {\pi}{15}\zeta(3)\cot\left(\frac {\pi}{5}\right)-\frac {\zeta(3)}{4}$

Using any trigonometric table or Wolfram alpha you can find that $\cot\left(\frac {\pi}{5}\right)=\sqrt{1+\frac {2}{\sqrt 5}}$ So we get $\displaystyle \xi=\frac {\pi}{15}\zeta(3)\sqrt{1+\frac {2}{\sqrt 5}}-\frac {\zeta(3)}{4}$

Hence the answer $\boxed {30}$