Answer
$5^{1/3}\approx1.709$
Work Step by Step
Given $$5^{1/3} $$
Let $$f(x) =x^3-5 $$
and suppose that $x_0=2$, 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}-5}{3x_n^2}\\
&=\frac{2x_n^{3}+5}{3x_n^2}
\end{align*}
Hence
\begin{aligned}
x_{1}&=\frac{2x_0^{3}+5}{3x_0^2}=\frac{2( 2)^{3}+5}{3(2) ^2} \approx 3.083 \\
x_{2}&=\frac{2x_1^{3}+5}{3x_1^2}= \frac{2( 3.083)^{3}+5}{3(3.083) ^2} \approx 2.230\\
x_{3}&=\frac{2x_2^{3}+5}{3x_2^2} =\frac{2(2.230)^{3}+5}{3(2.230) ^2} \approx1.821\\
x_{4}&=\frac{2x_3^{3}+5}{3x_3^2} =\frac{2(1.821)^{3}+5}{3(1.821) ^2} \approx1.716\\
x_{5}&=\frac{2x_4^{3}+5}{3x_4^2} =\frac{2(1.716)^{3}+5}{3(1.716) ^2} \approx1.709\\
x_{6}&=\frac{2x_5^{3}+5}{3x_5^2} =\frac{2(1.709)^{3}+5}{3(1.709) ^2} \approx1.709\\
\end{aligned}
Since $x_5=x_6 $, then $5^{1/3}\approx1.709$
By calculator we get
$$5^{1/3}=1.7099$$
