First, you can factor out a common factor of 2. Then, 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\$ .

$\displaystyle{128}+{2}{{d}}^{{3}}={2}{\left({64}+{{d}}^{{3}}\right)}={2}{\left({{4}}^{{3}}+{{d}}^{{3}}\right)}={2}{\left({4}+{d}\right)}{\left({16}-{4}{d}+{{d}}^{{2}}\right)}\usepackage{cmbright}\Huge\$