Correct. Using the rules of exponents, you find that the expression $\displaystyle\frac{{{\left({{a}}^{{-{{3}}}}{{b}}^{{2}}{{c}}^{{4}}\right)}}^{{2}}}{{{\left({a}{{b}}^{{-{{3}}}}{{c}}^{{2}}\right)}}^{{3}}}\usepackage{cmbright}\Huge\$ simplifies to $\displaystyle\frac{{{{a}}^{{-{{6}}}}{{b}}^{{4}}{{c}}^{{8}}}}{{{{a}}^{{3}}{{b}}^{{-{{9}}}}{{c}}^{{6}}}}\usepackage{cmbright}\Huge\$ , which is the same as $\displaystyle{{a}}^{{-{6}-{3}}}{{b}}^{{{4}-{\left(-{9}\right)}}}{{c}}^{{{8}-{6}}}={{a}}^{{-{{9}}}}{{b}}^{{13}}{{c}}^{{2}}=\frac{{{{b}}^{{13}}{{c}}^{{2}}}}{{{a}}^{{9}}}\usepackage{cmbright}\Huge\$ .