Answer
$$\root 3 \of 6 \approx 1.817120593$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = {x^3} - 6,{\text{ so }}f'\left( x \right) = 3{x^2} \cr
& {x^3} - 6 = 0 \cr
& x = \root 3 \of 6 \cr
& {\text{Using the Newton's Method }}{x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}{\text{ we have}} \cr
& {\text{Then evaluating }}f\left( {{x_n}} \right){\text{ and }}f'\left( {{x_n}} \right) \cr
& {x_{n + 1}} = {x_n} - \frac{{x_n^3 - 6}}{{3x_n^2}} \cr
& \left( 1 \right)\left( 1 \right)\left( 1 \right) = 1{\text{ and }}\left( 2 \right)\left( 2 \right)\left( 2 \right) = 8 \cr
& {\text{Taking }}{x_1} = 1 \cr
& {x_1} = 1 \cr
& {x_{1 + 1}} = {x_2} = 1 - \frac{{{{\left( 1 \right)}^3} - 6}}{{3{{\left( 1 \right)}^2}}} \approx 2.666666667 \cr
& {x_{2 + 1}} = {x_3} = 2.666666667 - \frac{{{{\left( {2.666666667} \right)}^3} - 6}}{{3{{\left( {2.666666667} \right)}^2}}} = 2.059027778 \cr
& {x_{3 + 1}} = {x_4} = 2.059027778 - \frac{{{{\left( {2.059027778} \right)}^3} - 6}}{{3{{\left( {2.059027778} \right)}^2}}} = 1.8444428315 \cr
& {x_{4 + 1}} = {x_5} = 1.8444428315 - \frac{{{{\left( {1.8444428315} \right)}^3} - 6}}{{3{{\left( {1.8444428315} \right)}^2}}} = 1.817522903 \cr
& {x_{5 + 1}} = {x_6} = 1.817522903 - \frac{{{{\left( {1.817522903} \right)}^3} - 6}}{{3{{\left( {1.817522903} \right)}^2}}} = 1.817120682 \cr
& {x_{6 + 1}} = {x_7} = 1.817120682 - \frac{{{{\left( {1.817120682} \right)}^3} - 6}}{{3{{\left( {1.817120682} \right)}^2}}} = 1.817120593 \cr
& {x_{7 + 1}} = {x_8} = 1.817120593 - \frac{{{{\left( {1.817120593} \right)}^3} - 6}}{{3{{\left( {1.817120593} \right)}^2}}} = 1.817120593 \cr
& {\text{We obtain that }}{x_7} \approx {x_8} \cr
& {\text{Thus}}{\text{,}} \cr
& \root 3 \of 6 \approx 1.817120593 \cr} $$
