Correct. If $\displaystyle{3}-{\sqrt[{{3}}]{{{x}+{1}}}}={1}\usepackage{cmbright}\Huge\$ , then $\displaystyle-{\sqrt[{{3}}]{{{x}+{1}}}}=-{2}\usepackage{cmbright}\Huge\$ , so $\displaystyle{\sqrt[{{3}}]{{{x}+{1}}}}={2}\usepackage{cmbright}\Huge\$ , and therefore $\displaystyle{{\left({\sqrt[{{3}}]{{{x}+{1}}}}\right)}}^{{3}}={{\left({2}\right)}}^{{3}}\usepackage{cmbright}\Huge\$ . Then $\displaystyle{x}+{1}={8}\usepackage{cmbright}\Huge\$ , so $\displaystyle{x}={7}\usepackage{cmbright}\Huge\$ .