Compute $\sum\limits_{a=1}^{\infty} \sum\limits_{b=1}^{\infty} \sum\limits_{c=1}^{\infty}\frac{ab(3a+c)}{4^{a+b+c}((a+b)(b+c)(a+c)}$ Compute the following sum:
$$
\sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \sum_{c=1}^{\infty}\frac{ab(3a+c)}{4^{a+b+c}((a+b)(b+c)(a+c)}\
$$
I have no head or tail of how to even start this since the summation is interlinked.

The basic idea is that, the final value of the sum doesn't depend on the dummy indices, i.e. whether you choose $(a,b,c)$, $(x,y,z)$ or some other triplet - it simply does not matter. Take advantage of this, and swap the indices $(a,b,c)$ around. How many permutations of these $3$ exist? Exactly $3! = 6$ permutations. I leave it to you to write them down. Now that you know that these $6$ different ways of writing the sum mean exactly the same thing - we're good to go. Denote the sum by $S$, and add all these (six) different representations of the same sum together. Get a nice expression for $6S$, and thus find $S$. See the magic unfold!
$$6S = \sum_{a = 1}^{\infty} \sum_{b = 1}^{\infty} \sum_{c = 1}^{\infty} \frac{ab(3a+c) + ac(3a+b) + bc(3b+a) + ba(3b+c) + ca(3c+b) + cb(3c+a)}{4^{a+b+c} (a+b)(b+c)(c+a)}$$
$$6S = \sum_{a = 1}^{\infty} \sum_{b = 1}^{\infty} \sum_{c = 1}^{\infty} \frac{3a^2b+3a^2c+3b^2c+3b^2a+3c^2a+3c^2b+6abc}{4^{a+b+c} (a+b)(b+c)(c+a)}$$
$$6S = \sum_{a = 1}^{\infty} \sum_{b = 1}^{\infty} \sum_{c = 1}^{\infty} \frac{3(a+b)(b+c)(c+a)}{4^{a+b+c} (a+b)(b+c)(c+a)}$$
$$S = \frac12\sum_{a = 1}^{\infty} \sum_{b = 1}^{\infty} \sum_{c = 1}^{\infty} \frac1{4^{a+b+c}} = \frac12\left(\sum_{a=1}^{\infty}\frac1{4^a}\right)^3 = \frac12\left(\frac13\right)^3 = \boxed{\frac1{54}}$$

If you want to practice this trick that I just mentioned, try to compute the following sum in a similar fashion (it has definitely been asked on this site before):
$$\sum_{m=1}^\infty\sum_{n=1}^\infty\dfrac{m^2n}{3^m(n3^m+m3^n)} = \frac{9}{32}$$
