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.