Incorrect. A binomial in the form $\displaystyle{{a}}^{{3}}+{{b}}^{{3}}\usepackage{cmbright}\Huge\$ can be factored as $\displaystyle{\left({a}+{b}\right)}{\left({{a}}^{{2}}-{a}{b}+{{b}}^{{2}}\right)}\usepackage{cmbright}\Huge\$ . Since $\displaystyle{8}{{x}}^{{3}}+{27}{{y}}^{{3}}={{\left({2}{x}\right)}}^{{3}}+{{\left({3}{y}\right)}}^{{3}}\usepackage{cmbright}\Huge\$ , $\displaystyle{a}={2}{x}\usepackage{cmbright}\Huge\$ and $\displaystyle{b}={3}{y}\usepackage{cmbright}\Huge\$ . The correct answer is $\displaystyle{6}{x}{y}\usepackage{cmbright}\Huge\$ .