Astonishing result

$i , j , k$ can be summed independently.

The summation corresponding to each of them is just

$\zeta(2)= \dfrac{\pi^2}{6}$

Thus, the required expression is just

$(\zeta(2))^3$ .

Thus, $b=216$ and $a=6$ .

It might look at first sight a huge mind boggling problem but if you have Basel's series in mind the problem may be solved at a sight. The problem is:- $\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{k=1}^{\infty} \dfrac{1}{(ijk)^{2}}=\dfrac{\pi ^{a}}{b}$ We can take $i$ and $j$ out of the first summation as being constant for summation over first symbol. $=\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \dfrac{1}{(ij)^{2}} \sum_{k=1}^{\infty} \dfrac{1}{k^{2}}$ From basel's formula:- $\sum_{i=1}^{\infty} \dfrac{1}{k^{2}} = \dfrac{\pi^{2}}{6}$ So our summation becomes:- $=\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \dfrac{1}{(ij)^{2}} \dfrac{\pi^{2}}{6}$ Repeating the same process $2$ more times we get the summation:- $=\dfrac{\pi^{6}}{216}$

why is it astonishing????