If n is the number of irrational terms in the expansion of  ${(3^{1/4}+5^{1/8})}^{60},$  then  $(n-1)$  is divisible by

${(3^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}+5^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$8$}\right.})}^{60} {}^{60}C_r{\left(3^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\right)}^{60-r}{\left(5^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}\right)}^r {}^{60}C_r{\left(3\right)}^{\frac{60-r}{4}}{5}^{\raisebox{1ex}{$r$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}$

For rational terms. 

  $\frac{r}{8}=k: 0\le r\le 60 0\le 8k\le 60$

$$\begin{gathered} 0\le 8k\le 60 0\le k\le \frac{60}{8} \\ 0\le k\le 7.5 k=0,1,2,3,4,5,6,7 \end{gathered}$$

$\frac{60-8k}{4}$  is always divisible by 4 for al value of k.
 Total rational terms  8
 Total terms 61
 Irrational terms  53
  $n-1=53-1=52$
 52 is divisible by 26.