$$\left(1+2^{\frac{1}{3}}+3^{\frac{1}{2}}\right)^6$$

$$ = {{\left| \!{\underline {\, 6 \,}} \right. } \over {\left| \!{\underline {\, {{r_1}} \,}} \right. \left| \!{\underline {\, {{r_2}} \,}} \right. \left| \!{\underline {\, {{r_3}} \,}} \right. }}{(1)^{{r_1}}}{(2)^{{{{r_2}} \over 3}}}{(3)^{{{{r_3}} \over 2}}}$$

$$\begin{array}{l|l|l} \mathrm{r}_1 & \mathrm{r}_2 & \mathrm{r}_3 \\ \hline 6 & 0 & 0 \\ 4 & 0 & 2 \\ 2 & 0 & 4 \\ 0 & 0 & 6 \\ \hline 3 & 3 & 0 \\ 1 & 3 & 2 \\ \hline 0 & 6 & 0 \\ \hline \end{array}$$

$$ = {{\left| \!{\underline {\, 6 \,}} \right. } \over {\left| \!{\underline {\, 6 \,}} \right. \left| \!{\underline {\, 0 \,}} \right. \left| \!{\underline {\, 0 \,}} \right. }} + {{\left| \!{\underline {\, 6 \,}} \right. } \over {\left| \!{\underline {\, 4 \,}} \right. \left| \!{\underline {\, 0 \,}} \right. \left| \!{\underline {\, 2 \,}} \right. }}(3) + {{\left| \!{\underline {\, 6 \,}} \right. } \over {\left| \!{\underline {\, 2 \,}} \right. \left| \!{\underline {\, 0 \,}} \right. \left| \!{\underline {\, 4 \,}} \right. }}{(3)^2} + {{\left| \!{\underline {\, 6 \,}} \right. } \over {\left| \!{\underline {\, 0 \,}} \right. \left| \!{\underline {\, 0 \,}} \right. \left| \!{\underline {\, 6 \,}} \right. }}{(3)^3} + {{\left| \!{\underline {\, 6 \,}} \right. } \over {\left| \!{\underline {\, 3 \,}} \right. \left| \!{\underline {\, 3 \,}} \right. \left| \!{\underline {\, 0 \,}} \right. }}(2) + {{\left| \!{\underline {\, 6 \,}} \right. } \over {\left| \!{\underline {\, 1 \,}} \right. \left| \!{\underline {\, 3 \,}} \right. \left| \!{\underline {\, 2 \,}} \right. }}{(2)^1}{(3)^1} + {{\left| \!{\underline {\, 6 \,}} \right. } \over {\left| \!{\underline {\, 0 \,}} \right. \left| \!{\underline {\, 6 \,}} \right. \left| \!{\underline {\, 0 \,}} \right. }}{(2)^2}$$

$=1+45+135+27+40+360+4=612$