We invert the given equation for \(n\) and write

\[
\mu_{0}=\frac{4 \pi a \hbar^{2} n}{m}\left[1+\frac{32}{3 \pi^{1 / 2}}\left(n a^{3}ight)^{1 / 2}+\ldotsight]
\]

Substituting this into the given expressions for \(E_{0}\) and \(P_{0}\), we get

\[
\begin{aligned}
\frac{E_{0}}{V} & =\frac{2 \pi a \hbar^{2} n^{2}}{m}\left[1+\frac{32}{3 \pi^{1 / 2}}\left(n a^{3}ight)^{1 / 2}+\ldotsight]^{2}\left[1-\frac{64}{5 \pi^{1 / 2}}\left(n a^{3}ight)^{1 / 2}+\ldotsight] \\
& =\frac{2 \pi a \hbar^{2} n^{2}}{m}\left[1+\frac{128}{15 \pi^{1 / 2}}\left(n a^{3}ight)^{1 / 2}+\ldotsight], \text { and } \\
P_{0} & =\frac{2 \pi a \hbar^{2} n^{2}}{m}\left[1+\frac{32}{3 \pi^{1 / 2}}\left(n a^{3}ight)^{1 / 2}+\ldotsight]^{2}\left[1-\frac{128}{15 \pi^{1 / 2}}\left(n a^{3}ight)^{1 / 2}+\ldotsight] \\
& =\frac{2 \pi a \hbar^{2} n^{2}}{m}\left[1+\frac{64}{5 \pi^{1 / 2}}\left(n a^{3}ight)^{1 / 2}+\ldotsight],
\end{aligned}
\]

in complete agreement with eqns. (11.3.16 and 17).