Novelty seems here

There need to be a subtle understanding that $a,b$ and $c$ are just variables and we can change the names of the variable and it won't change the value of the sum.

Also, it doesn't matter how we are summing up these terms as the summation sign can be interchanged.

Considering these two facts, we can make 6 total permutations of the above equation by exchanging the variables.

Adding all these together we will get,

$\displaystyle\sum_{a=1}^\infty \sum_{b=1}^\infty \sum_{c=1}^\infty \dfrac{\beta(a^{2}b+a^{2}c+b^{2}a+b^{2}c+c^{2}a+c^{2}b)+6abc}{\alpha^{a+b+c}(a+b)(b+c)(c+a)}=\dfrac{1}{9}$

I can't prove why and my solution is just what I did, is to take $\beta=3$ , so that we can factor the numerator and cancel it out with the term in the denominator. We'll be left with,

$\displaystyle\sum_{a=1}^\infty \sum_{b=1}^\infty \sum_{c=1}^\infty \dfrac{3}{\alpha^{a+b+c}}=\dfrac{1}{9}$

Now we can solve it easily and the simplified version looks like:

$\dfrac{3}{(\alpha-1)^{3}}=\dfrac{1}{9}$

which gives $\alpha=4$ .

Thus $\alpha^{\beta}=64$ .

P.S. Please suggest me a reasonable motivation to get the value of $\beta$ if someone finds it.