If the initial guess of the solution of
\[
1+x-x^3=0
\]
is \(x_0=1\), then the \(2^{\text{nd}}\) approximate root of
\[
1+x-x^3=0
\]
using Newton-Raphson method is
Show Hint
For Newton-Raphson problems, carefully compute
\[
f(x_n)\quad\text{and}\quad f'(x_n)
\]
before substituting into
\[
x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}.
\]
Arithmetic mistakes are the most common source of errors.