Answer
$$\eqalign{
& {x_{n + 1}} = {x_n} - \frac{{x_n^3 - 2}}{{3x_n^2}} \cr
& {x_1} = 1.5 \cr
& {x_2} = 1.296296 \cr} $$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = {x^3} - 2,{\text{ and }}{x_0} = 2 \cr
& {\text{Using the Newton's Method }}{x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}} \cr
& f'\left( x \right) = 3{x^2} \cr
& {\text{Then evaluating }}f\left( {{x_n}} \right){\text{ and }}f'\left( {{x_n}} \right) \cr
& {x_{n + 1}} = {x_n} - \frac{{x_n^3 - 2}}{{3x_n^2}} \cr
& \cr
& {\text{Taking }}{x_0} = 2 \cr
& {x_0} = 2 \cr
& {x_{0 + 1}} = {x_1} = 2 - \frac{{{{\left( 2 \right)}^2} - 2}}{{3\left( 2 \right)}} = 1.5 \cr
& {x_{1 + 1}} = {x_2} = 1.5 - \frac{{{{\left( {1.5} \right)}^2} - 2}}{{3\left( {1.5} \right)}} = 1.296296 \cr
& \cr
& {\text{Thus}}{\text{, we obtain}} \cr
& {x_{n + 1}} = {x_n} - \frac{{x_n^3 - 2}}{{3x_n^2}} \cr
& {x_1} = 1.5 \cr
& {x_2} = 1.296296 \cr} $$
