Sum of rational terms in ${\left(\sqrt{2}+\sqrt[5]{3}\right)}^{10}$ is

${\left(\sqrt{2}+\sqrt[5]{3}\right)}^{10}$

G.T $={}^{10}{\mathrm{C}}_{\mathrm{r}} {\left(2^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}^{10-\mathrm{r}} {\left({\left(3\right)}^{\frac{1}{5}}\right)}^{\mathrm{r}}$

       $={}^{10}{\mathrm{C}}_{\mathrm{r}} 2\frac{10-\mathrm{r}}{2} 3\frac{\mathrm{r}}{5}$

$\mathrm{r}=0, 5, 10, \frac{10-\mathrm{r}}{2} \in$ Integer

$\Rightarrow \mathrm{r}=0, 10$ are possible.

Sum$\begin{array}{rcl} & =& {}^{10}{\mathrm{C}}_0 {\left(\sqrt{2}\right)}^{10} {\left({\left(3\right)}^{\frac{1}{5}}\right)}^0+{}^{10}{\mathrm{C}}_{10} {\left(\sqrt{2}\right)}^0 {\left(3^{\frac{1}{5}}\right)}^{10}\\ & =& 2^5+3^2=41\end{array}$