Answer
\[\boxed{\begin{array}{*{20}{c}}
n&{{x_n}} \\
0&{ - 2.000000} \\
1&{ - 1.625000} \\
2&{ - 1.485786} \\
3&{ - 1.465956} \\
4&{ - 1.465571} \\
5&{ - 1.465571} \\
6&{ - 1.465571} \\
7&{ - 1.465571} \\
8&{ - 1.465571} \\
9&{ - 1.465571} \\
{10}&{ - 1.465571}
\end{array}}\]
Work Step by Step
\[\begin{gathered}
{\text{Let }}f\left( x \right) = {x^3} + {x^2} + 1,{\text{ and }}{x_0} = - 2 \hfill \\
{\text{Using the Newton's Method }}{x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}} \hfill \\
f'\left( x \right) = 3{x^2} + 2x \hfill \\
{\text{Then evaluating }}f\left( {{x_n}} \right){\text{ and }}f'\left( {{x_n}} \right) \hfill \\
{x_{n + 1}} = {x_n} - \frac{{x_n^3 + x_n^2 + 1}}{{3x_n^2 + 2{x_n}}} \hfill \\
\hfill \\
{\text{Taking }}{x_0} = - 2 \hfill \\
{x_0} = - 2 \hfill \\
{x_{0 + 1}} = {x_1} = - 2 - \frac{{{{\left( { - 2} \right)}^3} + {{\left( { - 2} \right)}^2} + 1}}{{3{{\left( { - 2} \right)}^2} + 2\left( { - 2} \right)}} = - 1.625000 \hfill \\
{x_{1 + 1}} = {x_2} = - 1.625000 - \frac{{{{\left( { - 1.625000} \right)}^3} + {{\left( { - 1.625000} \right)}^2} + 1}}{{3{{\left( { - 1.625000} \right)}^2} + 2\left( { - 1.625000} \right)}} \approx - 1.485786 \hfill \\
{x_{1 + 2}} = {x_3} = - 1.485786 - \frac{{{{\left( { - 1.485786} \right)}^3} + {{\left( { - 1.485786} \right)}^2} + 1}}{{3{{\left( { - 1.485786} \right)}^2} + 2\left( { - 1.485786} \right)}} \approx - 1.465956 \hfill \\
{x_{1 + 3}} = {x_4} = - 1.465956 - \frac{{{{\left( { - 1.465956} \right)}^3} + {{\left( { - 1.465956} \right)}^2} + 1}}{{3{{\left( { - 1.465956} \right)}^2} + 2\left( { - 1.465956} \right)}} \approx - 1.465571 \hfill \\
{x_{1 + 4}} = {x_5} = - 1.465571 - \frac{{{{\left( { - 1.465571} \right)}^3} + {{\left( { - 1.465571} \right)}^2} + 1}}{{3{{\left( { - 1.465571} \right)}^2} + 2\left( { - 1.465571} \right)}} \approx - 1.465571 \hfill \\
{x_{1 + 5}} = {x_6} = - 1.465571 - \frac{{{{\left( { - 1.465571} \right)}^3} + {{\left( { - 1.465571} \right)}^2} + 1}}{{3{{\left( { - 1.465571} \right)}^2} + 2\left( { - 1.465571} \right)}} \approx - 1.465571 \hfill \\
{x_{1 + 6}} = {x_7} = - 1.465571 - \frac{{{{\left( { - 1.465571} \right)}^3} + {{\left( { - 1.465571} \right)}^2} + 1}}{{3{{\left( { - 1.465571} \right)}^2} + 2\left( { - 1.465571} \right)}} \approx - 1.465571 \hfill \\
{x_{1 + 7}} = {x_8} = - 1.465571 - \frac{{{{\left( { - 1.465571} \right)}^3} + {{\left( { - 1.465571} \right)}^2} + 1}}{{3{{\left( { - 1.465571} \right)}^2} + 2\left( { - 1.465571} \right)}} \approx - 1.465571 \hfill \\
{x_{1 + 8}} = {x_9} = - 1.465571 - \frac{{{{\left( { - 1.465571} \right)}^3} + {{\left( { - 1.465571} \right)}^2} + 1}}{{3{{\left( { - 1.465571} \right)}^2} + 2\left( { - 1.465571} \right)}} \approx - 1.465571 \hfill \\
{x_{1 + 9}} = {x_{10}} = - 1.465571 - \frac{{{{\left( { - 1.465571} \right)}^3} + {{\left( { - 1.465571} \right)}^2} + 1}}{{3{{\left( { - 1.465571} \right)}^2} + 2\left( { - 1.465571} \right)}} \approx - 1.465571 \hfill \\
\hfill \\
{\text{Thus}}{\text{, we obtain}} \hfill \\
\boxed{\begin{array}{*{20}{c}}
n&{{x_n}} \\
0&{ - 2.000000} \\
1&{ - 1.625000} \\
2&{ - 1.485786} \\
3&{ - 1.465956} \\
4&{ - 1.465571} \\
5&{ - 1.465571} \\
6&{ - 1.465571} \\
7&{ - 1.465571} \\
8&{ - 1.465571} \\
9&{ - 1.465571} \\
{10}&{ - 1.465571}
\end{array}} \hfill \\
\end{gathered} \]
