#### Answer

\begin{aligned}
x_{1}& \approx 2.166667 \\
x_{2}& \approx 2.154504 \\
x_{3}& \approx 2.154435
\end{aligned}

#### Work Step by Step

Given $$ f(x)=x^{3}-10, \quad x_{0}=2 $$
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}-10}{3x_n^2}\\
&=\frac{2 x^{3}+10}{3 x^{2} n}
\end{align*}
Hence
\begin{aligned}
x_{1}&=\frac{2 x_{0}^{3}+10}{3 x_{0}^{2}}= \frac{2 \cdot 2^{3}+10}{3 \cdot 2^{2}} \approx 2.166667 \\
x_{2}&=\frac{2 x_{1}^{3}+10}{3 x_{1}^{2}}=\frac{\frac{25}{4}+6}{5} \approx 2.154504 \\
x_{3}&=\frac{2 x_{2}^{3}+10}{3 x_{2}^{2}} =\frac{2(2.154504)^{3}+10}{3 (2.154504)^{2}} \approx 2.154435
\end{aligned}