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$$
{\mathcal  {S}}\equiv {\biggl [}-{\frac  {b_{0}}{2}}+{\mathcal  {D}}^{{1/2}}{\biggr ]}^{{1/3}}={\biggl [}{\frac  {a_{1}^{2}}{2}}+{\mathcal  {D}}^{{1/2}}{\biggr ]}^{{1/3}}={\biggl [}{\frac  {a_{1}^{2}}{2}}{\biggr ]}^{{1/3}}{\biggl \{}1+{\biggl [}1+{\biggl (}{\frac  {8a_{0}}{3a_{1}^{2}}}{\biggr )}^{2}{\biggr ]}^{{1/2}}{\biggr \}}^{{1/3}},
$$
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