Answer
$$x \approx 1.769292354$$
Work Step by Step
$$\eqalign{
& {x^3} - 2x - 2 = 0 \cr
& {\text{Let }}f\left( x \right) = {x^3} - 2x - 2,{\text{ so }}f'\left( x \right) = 3{x^2} - 2 \cr
& {\text{Using the Newton's Method }}{x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}{\text{ we have}} \cr
& {x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}} \cr
& {\text{Then evaluating }}f\left( {{x_n}} \right){\text{ and }}f'\left( {{x_n}} \right) \cr
& {x_{n + 1}} = {x_n} - \frac{{x_n^3 - 2{x_n} - 2}}{{3x_n^2 - 2}} \cr
& \cr
& {\text{Evaluating }}f\left( 1 \right){\text{ and }}f\left( 2 \right) \cr
& f\left( 1 \right) = {\left( 1 \right)^3} - 2\left( 1 \right) - 2 = - 3 \cr
& f\left( 2 \right) = {\left( 2 \right)^3} - 2\left( 2 \right) - 2 = 2 \cr
& {\text{Then}}{\text{, Thus this solution lies between 1 and 2}} \cr
& {\text{We will use }}{x_1}{\text{ = 1}}{\text{.5 as our first approximation}} \cr
& {x_1} = 1.5 \cr
& {x_{1 + 1}} = {x_2} = 1.5 - \frac{{{{\left( {1.5} \right)}^3} - 2\left( {1.5} \right) - 2}}{{3{{\left( {1.5} \right)}^2} - 2}} \approx 1.842105263 \cr
& {x_{2 + 1}} = \cr
& {x_3} = 1.842105263 - \frac{{{{\left( {1.842105263} \right)}^3} - 2\left( {1.842105263} \right) - 2}}{{3{{\left( {1.842105263} \right)}^2} - 2}} \approx 1.77282692 \cr
& {x_{3 + 1}} = \cr
& {x_4} = 1.77282692 - \frac{{{{\left( {1.77282692} \right)}^3} - 2\left( {1.77282692} \right) - 2}}{{3{{\left( {1.77282692} \right)}^2} - 2}} \approx 1.769301293 \cr
& {x_{4 + 1}} = \cr
& {x_5} = 1.769301293 - \frac{{{{\left( {1.769301293} \right)}^3} - 2\left( {1.769301293} \right) - 2}}{{3{{\left( {1.769301293} \right)}^2} - 2}} \approx 1.769292354 \cr
& {x_{5 + 1}} = \cr
& {x_6} = 1.769292354 - \frac{{{{\left( {1.769292354} \right)}^3} - 2\left( {1.769292354} \right) - 2}}{{3{{\left( {1.769292354} \right)}^2} - 2}} \approx 1.769292354 \cr
& {\text{We obtain that }}{x_5} \approx {x_6} \cr
& {\text{Thus}}{\text{, the real solution is:}} \cr
& x \approx 1.769292354 \cr} $$
