Correct. First, notice that $\displaystyle{{8}}^{{\frac{{1}}{{3}}}}={2}\usepackage{cmbright}\Huge\$ , since $\displaystyle{{2}}^{{3}}={8}\usepackage{cmbright}\Huge\$ . Therefore, $\displaystyle{{8}}^{{\frac{{2}}{{3}}}}={{\left({{8}}^{{\frac{{1}}{{3}}}}\right)}}^{{2}}={{2}}^{{2}}={4}\usepackage{cmbright}\Huge\$ .