Since $\displaystyle{2}{x}{{y}}^{{2}}{\left({6}{x}{y}-{5}\right)}={12}{{x}}^{{2}}{{y}}^{{3}}-{10}{x}{{y}}^{{2}}\usepackage{cmbright}\Huge\$ but $\displaystyle{2}{x}{{y}}^{{2}}{\left({x}{y}-{1}\right)}={2}{{x}}^{{2}}{{y}}^{{3}}-{2}{x}{{y}}^{{2}}\ne{12}{{x}}^{{2}}{{y}}^{{3}}-{10}{x}{{y}}^{{2}}\usepackage{cmbright}\Huge\$ , this is the correct answer.