a) \({{{{\left( {\root 4 \of {{a^3}{b^2}} } \right)}^4}} \over {\root 3 \of {\sqrt {{a^{12}}{b^6}} } }} = {{{a^3}{b^2}} \over {\root 6 \of {{a^{12}}{b^6}} }} = {{{a^3}{b^2}} \over {{a^2}b}} = ab\)
b) \({{{a^{{1 \over 3}}} - {a^{{7 \over 3}}}} \over {{a^{{1 \over 3}}} - {a^{{4 \over 3}}}}} - {{{a^{ - {1 \over 3}}} - {a^{{5 \over 3}}}} \over {{a^{{2 \over 3}}} + {a^{ - {1 \over 3}}}}} = {{{a^{{1 \over 3}}}\left( {1 - {a^2}} \right)} \over {{a^{{1 \over 3}}}\left( {1-a} \right)}} - {{{a^{ - {1 \over 3}}}\left( {1 - {a^2}} \right)} \over {{a^{ - {1 \over 3}}}\left( {a + 1} \right)}} = \left( {1 + a} \right) - \left( {1 - a} \right) = 2a.\)
