Incorrect. $\displaystyle{{16}}^{{\frac{{1}}{{2}}}}={4}\usepackage{cmbright}\Huge\$ , so $\displaystyle{{16}}^{{-\frac{{1}}{{2}}}}=\frac{{1}}{{4}}\usepackage{cmbright}\Huge\$ . Use the fact that $\displaystyle{{16}}^{{\frac{{3}}{{2}}}}={{\left({{16}}^{{\frac{{1}}{{2}}}}\right)}}^{{3}}\usepackage{cmbright}\Huge\$ and $\displaystyle{{16}}^{{-\frac{{3}}{{2}}}}=\frac{{1}}{{{{16}}^{{\frac{{3}}{{2}}}}}}\usepackage{cmbright}\Huge\$ to simplify the expression.