The number of integral terms in the expansion of $(\sqrt{3}+\sqrt[8]{5}{)}^{256}$ is:

The general term in the expansion is:
$$T_{r+1}={}^{256}{C}_r(3^{1/2}{)}^{256-r}(5^{1/8}{)}^r={}^{256}{C}_r{3}^{\frac{256-r}{2}}{5}^{\frac{r}{8}}$$
For the term to be an integer, the powers of 3 and 5 must be integers.$\frac{256-r}{2}$ is an integer $⟹r$ is even.$\frac{r}{8}$ is an integer $⟹r$ is a multiple of 8.Thus, $r$ must be a multiple of 8.Possible values for $r$ are $0,8,16,\dots ,256$.This is an A.P. with $a=0,d=8,l=256$.Number of terms $n=\frac{256-0}{8}+1=32+1=33$.