3 Three's

\begin{aligned} \sum_{i = 0}^\infty \sum_{\substack{j = 0\\ i \ne j \ne k}}^\infty \sum_{k = 0}^\infty \frac 1{3^i 3^j 3^k} & = \sum_{i = 0}^\infty \sum_{j = 0}^\infty \sum_{k = 0}^\infty \frac 1{3^i 3^j 3^k} - \underbrace{\sum_{i = j = 0}^\infty \sum_{k = 0}^\infty \frac 1{3^{2i} 3^k}}_{S_1} - \underbrace{\sum_{i = 0}^\infty \sum_{j = k = 0}^\infty \frac 1{3^i 3^{2j}}}_{S_1} - \underbrace{\sum_{k=i = 0}^\infty \sum_{j= 0}^\infty \frac 1{3^j 3^{2k}}}_{S_1} + 2 \underbrace{\sum_{i=j=k=0}^\infty \frac 1{3^{3i}}}_{S_2} & \small \color{#3D99F6} \text{Since every }S_1 \text{ has an }S_2, \\ & = \sum_{i = 0}^\infty \frac 1{3^i} \sum_{j = 0}^\infty \frac 1{3^j} \sum_{k = 0}^\infty \frac 1{3^k} - 3 \sum_{i = 0}^\infty \frac 1{3^i} \sum_{j = 0}^\infty \frac 1{9^j} + 2 \sum_{i = 0}^\infty \frac 1{3^i} \frac 1{27^i} & \small \color{#3D99F6} \text{we need to add back }2 S_2. \\ & = \left(\frac 1{1-\frac 13}\right)^3 - 3 \left(\frac 1{1-\frac 13} \right) \left(\frac 1{1-\frac 19} \right) + 2 \left(\frac 1{1-\frac 1{27}} \right) \\ & = \frac {27}8 - 3 \cdot \frac 32 \cdot \frac 98 + 2 \cdot \frac {27}{26} = \frac {81}{208} \end{aligned}

Therefore, $a+b = 81+208 = \boxed{289}$ .

Required cases = (All cases) - (i=j, j=k, i=k) + 2(i=j=k)

All cases means no conditions on i,j,k. The second term of the above expressions subtracted the case of i=j=k thrice. Since it only needs to be subtracted once, we add twice that case.

$\sum _{i=0}^{\infty }a^i =\frac {1}{1-a}$ for $a≤1$ .

Hence, All cases = $(\frac {1}{1-\frac {1}{3}})^3=(\frac {3}{2})^3=\frac {27}{8}$

If i=j, the sum becomes $\sum _{k=0} \sum _{i=0}\frac {1}{3^k9^i}=\frac {3}{2}×\frac {1}{1-\frac {1}{9}}=\frac {27}{16}$

If i=j=k, then the sum becomes $\sum _{i=0} \frac {1}{27^i}=\frac {27}{26}$

Put them in the very first equation to get the answer as $\frac {81}{208}$