$\displaystyle{{x}}^{{\frac{{8}}{{3}}}}{{y}}^{{-\frac{{4}}{{3}}}}={{\left({{x}}^{{8}}\right)}}^{{\frac{{1}}{{3}}}}{{\left({{y}}^{{-{{4}}}}\right)}}^{{\frac{{1}}{{3}}}}={{\left({{x}}^{{8}}{{y}}^{{-{4}}}\right)}}^{{\frac{{1}}{{3}}}}={\sqrt[{{3}}]{{{{x}}^{{8}}{{y}}^{{-{4}}}}}}={\sqrt[{{3}}]{{\frac{{{x}}^{{8}}}{{{y}}^{{4}}}}}}=\usepackage{cmbright}\Huge\$

$\displaystyle{\sqrt[{{3}}]{{\frac{{{{x}}^{{6}}{{x}}^{{2}}}}{{{{y}}^{{3}}{y}}}}}}={\sqrt[{{3}}]{{\frac{{{{\left({{x}}^{{2}}\right)}}^{{3}}{{x}}^{{2}}}}{{{{y}}^{{3}}{y}}}}}}=\frac{{{x}}^{{2}}}{{y}}{\sqrt[{{3}}]{{\frac{{{{x}}^{{2}}}}{{y}}}}}\usepackage{cmbright}\Huge\$ . Technically this is not in simplified form because there is a radical in the denominator. To clear it, multiply the inside of the radical by $\displaystyle\frac{{{y}}^{{2}}}{{{y}}^{{2}}}\usepackage{cmbright}\Huge\$ , and then factor out another $\displaystyle{\sqrt[{{3}}]{{{{y}}^{{3}}}}}\usepackage{cmbright}\Huge\$ from the denominator.