Answer
$2^{7/3}\approx5.039$
Work Step by Step
Given $$2^{7/3}= (2^7)^{1/3}= (128)^{1/3} $$
Let $$f(x) =x^3-128 $$
and suppose that $x_0=5$, then
Since
$$ x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}$$
Then
\begin{align*}
x_{n+1}&= x_n -\frac{x_n^3-128 }{3x_n^2}\\
&=\frac{2x_n^3+128 }{3x_n^2}
\end{align*}
Hence
\begin{aligned}
x_{1}&=\frac{2x_0^3+128 }{3x_0^2}=\frac{2(5)^3+128 }{3(5)^2} \approx 5.04 \\
x_{2}&=\frac{2x_1^3+128 }{3x_1^2}=\frac{2(5.04)^3+128 }{3(5.04)^2} \approx5.039\\
x_{3}&=\frac{2x_2^3+128 }{3x_2^2} =\frac{2(5.039)^3+128 }{3(5.039)^2} \approx5.039
\end{aligned}
Since $x_2=x_3 $, then $2^{7/3}\approx5.039$
By calculator we get
$$2^{7/3}=5.03968$$
