Correct. Using the rules of exponents, this expression can be simplified like this: $\displaystyle{{\left({{a}}^{{-{{12}}}}{{b}}^{{3}}{{c}}^{{-{{5}}}}\right)}}^{{-{{3}}}}\usepackage{cmbright}\Huge\$ = $\displaystyle{{a}}^{{-{12}{\left(-{3}\right)}}}{{b}}^{{{3}{\left(-{3}\right)}}}{{c}}^{{-{5}{\left(-{3}\right)}}}\usepackage{cmbright}\Huge\$ = $\displaystyle{{a}}^{{36}}{{b}}^{{-{{9}}}}{{c}}^{{15}}\usepackage{cmbright}\Huge\$ .