Answer
$$4.642$$
Work Step by Step
$$\eqalign{
& \root 3 \of {100} \cr
& {\text{We can note that }}\root 3 \of {100} {\text{ is a solution of the equation }}{x^3} - 100 = 0.{\text{ }} \cr
& {\text{Let }}f\left( x \right) = {x^3} - 100,\,\,\,\,\,\,f'\left( x \right) = 3{x^2} \cr
& {\text{Since 4 < }}\root 3 \of {100} < 5,{\text{ we can use }}{c_1} = 4{\text{ as the first approximation }} \cr
& {\text{Then using Newton's method we have}}: \cr
& {c_2} = {c_1} - \frac{{f\left( {{c_1}} \right)}}{{f'\left( {{c_1}} \right)}} = 4 - \frac{{{{\left( 4 \right)}^3} - 100}}{{3{{\left( 4 \right)}^2}}} = 4.75 \cr
& {c_3} = {c_2} - \frac{{f\left( {{c_2}} \right)}}{{f'\left( {{c_2}} \right)}} = 4.75 - \frac{{{{\left( {4.75} \right)}^3} - 100}}{{3{{\left( {4.75} \right)}^2}}} = 4.6440 \cr
& {\text{In the same way}} \cr
& {c_4} = 4.6440 - \frac{{{{\left( {4.6440} \right)}^3} - 100}}{{3{{\left( {4.6440} \right)}^2}}} = 4.6415 \cr
& {c_5} = 4.6415 - \frac{{{{\left( {4.6415} \right)}^3} - 100}}{{3{{\left( {4.6415} \right)}^2}}} = 4.64158 \cr
& {\text{Since }}{c_4} = {c_5} = 4.641,{\text{ to the nearest thousand}}{\text{, }}\root 3 \of {100} = 4.642 \cr} $$
