First, you can factor out a common factor of 4. 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{500}+{4}{{d}}^{{3}}={4}{\left({125}+{{d}}^{{3}}\right)}={4}{\left({{5}}^{{3}}+{{d}}^{{3}}\right)}={4}{\left({5}+{d}\right)}{\left({25}-{5}{d}+{{d}}^{{2}}\right)}\usepackage{cmbright}\Huge\$