By using $2\cdot 4^3-5^3=3$,obtain the formula
\begin{equation}
\label{eq:12.38}
\sqrt[3]{2}=\frac{5}{4}(1+\frac{1}{1\cdot 125}-\frac{2}{1\cdot
2\cdot (125)^2}+\frac{2\cdot 5}{1\cdot 2\cdot 3\cdot
(125)^3}-\frac{2\cdot 5\cdot 8}{1\cdot 2\cdot 3\cdot 4
(125)^4}+\cdots )
\end{equation}
Proof:
\begin{equation}
\label{eq:12.45}
\sqrt[3]{2}=\frac{5}{4}\sqrt[3]{1+\frac{3}{5^3}}
\end{equation}
Now we expande
\begin{equation}
\label{eq:12.49}
(1+x)^{\frac{1}{3}}
\end{equation}($|x|<1$)as
\begin{equation}
\label{eq:12.50}
1+\frac{\frac{1}{3}}{1!}x+\frac{\frac{1}{3}(\frac{1}{3}-1)}{2!}x^2+\frac{\frac{1}{3}(\frac{1}{3}-1)(\frac{1}{3}-2)}{3!}x^3+\cdots
\end{equation}
Simplify \ref{eq:12.50} as
\begin{equation}
\label{eq:12.55}
1+\frac{1}{3}x-\frac{1\cdot 2}{3^22!}x^2+\frac{1\cdot 2\cdot 5}{3^33!}x^3+\cdots
\end{equation}
Let $x=\frac{3}{5^3}$,then \ref{eq:12.55} becomes
\begin{equation}
\label{eq:1.05}
1+\frac{1}{5^3}-\frac{1\cdot 2}{2!5^3}+\frac{1\cdot 2\cdot 5}{3!5^3}+\cdots
\end{equation}
So \ref{eq:12.38} is verified.