Correct. If $\displaystyle{\sqrt[{{3}}]{{{x}}}}={4}\usepackage{cmbright}\Huge\$ , then $\displaystyle{{\left({\sqrt[{{3}}]{{{x}}}}\right)}}^{{3}}={{4}}^{{3}}={64}\usepackage{cmbright}\Huge\$ , so $\displaystyle{x}={64}\usepackage{cmbright}\Huge\$ .