Answer
$3^{-1/4}\approx0.759$
Work Step by Step
Given $$3^{-1/4}= (1/3)^{1/4}= $$
Let $$f(x) =x^4-1/3 $$
and suppose that $x_0=1$, 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^4-1/3 }{4x_n^3}\\
&=\frac{x_n^4+1/3 }{4x_n^3}
\end{align*}
Hence
\begin{aligned}
x_{1}&=\frac{x_0^4+1/3 }{4x_0^3}=\frac{(1)^4+1/3 }{4(1) ^3} \approx 0.833 \\
x_{2}&=\frac{x_1^4+1/3 }{4x_1^3}=\frac{(0.833)^4+1/3 }{4(0.833) ^3} \approx0.768\\
x_{3}&=\frac{x_2^4+1/3 }{4x_2^3} =\frac{(0.768)^4+1/3 }{4(0.768) ^3} \approx0.759\\
x_{4}&=\frac{x_3^4+1/3 }{4x_3^3} =\frac{(0.759)^4+1/3 }{4(0.759) ^3} \approx0.759
\end{aligned}
Since $x_3=x_4 $, then $3^{-1/4}\approx0.759$
By calculator we get
$$3^{-1/4}=0.759$$
