Simplify the fraction within the radical first: $\displaystyle{\sqrt[{{3}}]{{\frac{{{96}{{x}}^{{{5}-{1}}}{{y}}^{{{23}-{3}}}}}{{{{z}}^{{{14}-{11}}}}}}}}={\sqrt[{{3}}]{{\frac{{{96}{{x}}^{{4}}{{y}}^{{20}}}}{{{{z}}^{{3}}}}}}}\usepackage{cmbright}\Huge\$ . Then pull out cubed terms: $\displaystyle{\sqrt[{{3}}]{{\frac{{{12}\cdot{{2}}^{{3}}\cdot{{x}}^{{3}}\cdot{x}\cdot{{\left({{y}}^{{6}}\right)}}^{{3}}\cdot{{y}}^{{2}}}}{{{{z}}^{{3}}}}}}}=\frac{{{2}{x}{{y}}^{{6}}}}{{z}}{\sqrt[{{3}}]{{{12}{x}{{y}}^{{2}}}}}\usepackage{cmbright}\Huge\$ .