Correct. First, notice that $\displaystyle{{16}}^{{\frac{{1}}{{2}}}}={4}\usepackage{cmbright}\Huge\$ , so $\displaystyle{{16}}^{{-\frac{{1}}{{2}}}}=\frac{{1}}{{4}}\usepackage{cmbright}\Huge\$ . Therefore, $\displaystyle{{16}}^{{-\frac{{3}}{{2}}}}={{\left({{16}}^{{-\frac{{1}}{{2}}}}\right)}}^{{3}}={{\left(\frac{{1}}{{4}}\right)}}^{{3}}=\frac{{1}}{{{4}}^{{3}}}=\frac{{1}}{{64}}\usepackage{cmbright}\Huge\$ .