You may have forgotten that if n is odd, $\displaystyle{\sqrt[{{n}}]{{{{x}}^{{n}}}}}={x}\usepackage{cmbright}\Huge\$ , so $\displaystyle{\sqrt[{{3}}]{{-{8}{{x}}^{{3}}}}}={\sqrt[{{3}}]{{{{\left(-{1}\right)}}^{{3}}}}}\cdot{\sqrt[{{3}}]{{{{2}}^{{3}}}}}\cdot{\sqrt[{{3}}]{{{{x}}^{{3}}}}}=-{2}{x}\usepackage{cmbright}\Huge\$ . The simplified form does not have an absolute value symbol. The correct answer is: $\displaystyle\sqrt{{{36}{{x}}^{{2}}}}\usepackage{cmbright}\Huge\$