Correct. If $\displaystyle{\sqrt[{{3}}]{{{2}{x}-{1}}}}={\sqrt[{{3}}]{{{x}+{2}}}}\usepackage{cmbright}\Huge\$ , then $\displaystyle{{\left({\sqrt[{{3}}]{{{2}{x}-{1}}}}\right)}}^{{3}}={{\left({\sqrt[{{3}}]{{{x}+{2}}}}\right)}}^{{3}}\usepackage{cmbright}\Huge\$ , so $\displaystyle{2}{x}-{1}={x}+{2}\usepackage{cmbright}\Huge\$ , and therefore $\displaystyle{x}={3}\usepackage{cmbright}\Huge\$ .