Correct. Since $\displaystyle{V}={{x}}^{{3}}\usepackage{cmbright}\Huge\$ , you know that $\displaystyle{\sqrt[{{3}}]{{{V}}}}={\sqrt[{{3}}]{{{{x}}^{{3}}}}}={x}\usepackage{cmbright}\Huge\$ . Therefore, $\displaystyle{x}={\sqrt[{{3}}]{{{V}}}}\usepackage{cmbright}\Huge\$ .