Correct. Using the rules of exponents, this expression can be simplified like this: $\displaystyle\frac{{{\left({{a}}^{{8}}{{b}}^{{-{{3}}}}\right)}}^{{4}}}{{{\left({{a}}^{{3}}{{b}}^{{4}}{{c}}^{{-{{2}}}}\right)}}^{{5}}}=\frac{{{{a}}^{{{8}{\left({4}\right)}}}{{b}}^{{-{3}{\left({4}\right)}}}}}{{{{a}}^{{{3}{\left({5}\right)}}}{{b}}^{{{4}{\left({5}\right)}}}{{c}}^{{-{2}{\left({5}\right)}}}}}\usepackage{cmbright}\Huge\$ , which is the same as $\displaystyle\frac{{{{a}}^{{32}}{{b}}^{{-{{12}}}}}}{{{{a}}^{{15}}{{b}}^{{20}}{{c}}^{{-{{10}}}}}}={{a}}^{{{32}-{15}}}{{b}}^{{-{12}-{20}}}{{c}}^{{10}}={{a}}^{{17}}{{b}}^{{-{{32}}}}{{c}}^{{10}}\usepackage{cmbright}\Huge\$ .