Explanation

(b)
The general term in the expansion of ${{(5}^{\frac{1}{6}} + 2^{\frac{1}{8}})}^{100}$ is given by
$T_{r + 1} = \ ^{100}C_{r}\left( 5^{\frac{1}{6}} \right)^{100 - r}{(2^{\frac{1}{8}})}^{r}$
As 5 and 2 are relatively prime, $T_{r + 1}$ will be rational, if
$\frac{100 - r}{6}$ and $\frac{r}{8}$ are both integers $ie$, if $100 - r$ is a multiple of 6 and $r$ is a multiple of 8. As $0 \leq r \leq 100,$ multiples of 8 upto 100 and corresponding value of $100 - r$ are
$r = 0,\ 8,\ 16,\ 24,\ldots.,\ 88,\ 96$
$ie$, $100 - r = 100,\ 92,\ 84,\ 76,\ldots,\ 12,\ 4$
Out of $100 - r$, multiples of 6 are 84, 60, 36, 12
$\therefore$ There are four rational terms
Hence, number of irrational terms is $101 - 4 = 97$