If n is the number of irrational terms in the expansion of ( 3 ^ 1...

${\left({3}^{1 / 4} + {5}^{1 / 8}\right)}^{60}$

$= Cr 60 {\left({3}^{1 / 4}\right)}^{60 - r} . {\left({5}^{1 / 8}\right)}^{r}$

$= Cr 60 \left(3\right)\dfrac{60 - r}{4} . 5\dfrac{r}{8}$

For rational terms.

$\dfrac{r}{8} = k; 0 \le r \le 60$

$0 \le 8k \le 60$

$0 \le k \le \dfrac{60}{8}$

$0 \le k \le 7 . 5$

$k = 0, 1, 2, 3, 4, 5, 6, 7$

$\dfrac{60 - 8k}{4}$ is always divisible by $4$ for all value of $k$

Total rational terms $= 8$

Total terms $= 61$

irrational terms $= 53$

$n - 1 = 53 - 1 = 52$

$52$ is divisible by $26$.