Just count 4

$\left(5^{\frac{1}{8}}+2^{\frac{1}{6}} \right)^{100} = \sum _{n=0}^{100} {\begin{pmatrix} 100 \\ n \end{pmatrix} 5^{\frac{100-n}{8}}2^{\frac{n}{6}}}$

For the term to be rational, $8|(100-n)$ and $6|n$ . And there are $4$ such $n$ for $0 \le n \le 100$ .

$\begin{array} {cc} n & 100-n \\ 12 = 6\times 2 & 88 = 8 \times 11 \\ 36 = 6\times 6 & 64 = 8 \times 8 \\ 60 = 6\times 10 & 40 = 8 \times 5 \\ 84 = 6\times 14 & 16 = 8 \times 2 \end{array}$

Since there are a total of $101$ terms, the number of irrational terms is:

$101-4= \boxed{97}$