Answer
\begin{aligned}
x_{1} &= -0.75 \\
x_{2} & \approx-0.6860465 \\
x_{3} & \approx-0.6823396
\end{aligned}
Work Step by Step
Given $$ f(x)=x^{3}+x+1, \quad x_{0}=-1 $$
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}+x_n+1}{3x_n^2+1}\\
&=\frac{2 x_n^{3}-1}{3 x^{2}_ n+1}
\end{align*}
Hence
\begin{aligned}
x_{1}=& \frac{2 x_{0}^{3}-1}{3 x_{0}^{2}+1}=\frac{2(-1)^{3}-1}{3(-1)^{2}+1} =-0.75 \\
x_{2}=& \frac{2 x_{1}^{3}-1}{3 x_{1}^{2}+1}=\frac{2(-0.75)^{3}-1}{3(-0.75)^{2}+1} \approx-0.6860465 \\
x_{3}=&\frac{2 x_{2}^{3}-1}{3 x_{2}^{2}+1} =\frac{2(-0.6860465)^{3}-1}{3(-0.6860465)^{2}+1} \approx-0.6823396
\end{aligned}
