You may have correctly done the first step: $\displaystyle{16}{{x}}^{{3}}-{250}={2}{\left({8}{{x}}^{{3}}-{125}\right)}\usepackage{cmbright}\Huge\$ . Now $\displaystyle{8}{{x}}^{{3}}-{125}\usepackage{cmbright}\Huge\$ is in the form $\displaystyle{{a}}^{{3}}-{{b}}^{{3}}\usepackage{cmbright}\Huge\$ and 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}}={{\left({2}{x}\right)}}^{{3}}\usepackage{cmbright}\Huge\$ and $\displaystyle{125}={{5}}^{{3}}\usepackage{cmbright}\Huge\$ , you should have replaced a by $\displaystyle{2}{x}\usepackage{cmbright}\Huge\$ and b by 5. Whenever you factor, remember to check your answer by multiplying the factors back together.