Pulling out cubed terms, you find that $\displaystyle{\sqrt[{{3}}]{{{160}{{a}}^{{11}}{{b}}^{{43}}{{c}}^{{17}}}}}={\sqrt[{{3}}]{{{{2}}^{{3}}\cdot{20}\cdot{{\left({{a}}^{{3}}\right)}}^{{3}}\cdot{{a}}^{{2}}\cdot{{\left({{b}}^{{14}}\right)}}^{{3}}\cdot{b}\cdot{{\left({{c}}^{{5}}\right)}}^{{3}}\cdot{{c}}^{{2}}}}}\usepackage{cmbright}\Huge\$ . This expression simplifies to $\displaystyle{2}{{a}}^{{3}}{{b}}^{{14}}{{c}}^{{5}}{\sqrt[{{3}}]{{{20}{{a}}^{{2}}{b}{{c}}^{{2}}}}}\usepackage{cmbright}\Huge\$ .