Answer
$${P_4}\left( x \right) = - 1 + \frac{1}{3}x + \frac{1}{9}{x^2} + \frac{5}{{81}}{x^3} + \frac{{10}}{{243}}{x^4}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \root 3 \of {x - 1} \cr
& f\left( x \right) = {\left( {x - 1} \right)^{1/3}} \cr
& {\text{Use the definition of Taylor Polynomial of Degree }}n\,\,\,\left( {{\text{see page 629}}} \right) \cr
& {\text{Let }}f{\text{ be a function that can be differentiated }}n{\text{ times at 0}}{\text{. The Taylor }} \cr
& {\text{polynomial of degree }}n{\text{ for }}f{\text{ at 0 is }} \cr
& {P_n}\left( x \right) = f\left( 0 \right) + \frac{{{f^{\left( 1 \right)}}\left( 0 \right)}}{{1!}}x + \frac{{{f^{\left( 2 \right)}}\left( 0 \right)}}{{2!}}{x^2} + \cdots + \frac{{{f^{\left( n \right)}}\left( 0 \right)}}{{n!}}{x^n} = \sum\limits_{i = 0}^n {\frac{{{f^{\left( n \right)}}\left( 0 \right)}}{{i!}}} {x^i} \cr
& {\text{Find the Taylor polynomials of degree 4 at 0}}{\text{. }} \cr
& {\text{then }}n = 4. \cr
& {\text{The }}n{\text{ - th derivatives are}} \cr
& {f^{\left( 1 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ {{{\left( {x - 1} \right)}^{1/3}}} \right] = \frac{1}{3}{\left( {x - 1} \right)^{ - 2/3}} \cr
& {f^{\left( 2 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{3}{{\left( {x - 1} \right)}^{ - 2/3}}} \right] = - \frac{2}{9}{\left( {x - 1} \right)^{ - 5/3}} \cr
& {f^{\left( 3 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ { - \frac{2}{9}{{\left( {x - 1} \right)}^{ - 5/3}}} \right] = \frac{{10}}{{27}}{\left( {x - 1} \right)^{ - 8/3}} \cr
& {f^{\left( 4 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{10}}{{27}}{{\left( {x - 1} \right)}^{ - 8/3}}} \right] = - \frac{{80}}{{81}}{\left( {x - 1} \right)^{ - 11/3}} \cr
& {\text{evaluate }}f\left( 0 \right),{f^{\left( 1 \right)}}\left( 0 \right),{f^{\left( 2 \right)}}\left( 0 \right),{f^{\left( 3 \right)}}\left( 0 \right),{f^{\left( 4 \right)}}\left( 0 \right) \cr
& f\left( 0 \right) = {\left( {0 - 1} \right)^{1/3}} = - 1 \cr
& {f^{\left( 1 \right)}}\left( 0 \right) = \frac{1}{3}{\left( {0 - 1} \right)^{ - 2/3}} = \frac{1}{3} \cr
& {f^{\left( 2 \right)}}\left( 0 \right) = - \frac{2}{9}{\left( {0 - 1} \right)^{ - 5/3}} = \frac{2}{9} \cr
& {f^{\left( 3 \right)}}\left( 0 \right) = \frac{{10}}{{27}}{\left( {0 - 1} \right)^{ - 8/3}} = \frac{{10}}{{27}} \cr
& {f^{\left( 4 \right)}}\left( 0 \right) = - \frac{{80}}{{81}}{\left( {0 - 1} \right)^{ - 11/3}} = \frac{{80}}{{81}} \cr
& {\text{Replace the found values into the definition of Taylor Polynomial of Degree }}n \cr
& {\text{for }}n = 4 \cr
& {P_4}\left( x \right) = f\left( 0 \right) + \frac{{{f^{\left( 1 \right)}}\left( 0 \right)}}{{1!}}x + \frac{{{f^{\left( 2 \right)}}\left( 0 \right)}}{{2!}}{x^2} + \frac{{{f^{\left( 3 \right)}}\left( 0 \right)}}{{3!}}{x^3} + \frac{{{f^{\left( 4 \right)}}\left( 0 \right)}}{{4!}}{x^4} \cr
& {P_4}\left( x \right) = - 1 + \frac{{1/3}}{{1!}}x + \frac{{2/9}}{{2!}}{x^2} + \frac{{10/27}}{{3!}}{x^3} + \frac{{80/81}}{{4!}}{x^4} \cr
& {\text{simplifying}} \cr
& {P_4}\left( x \right) = - 1 + \frac{1}{3}x + \frac{1}{9}{x^2} + \frac{5}{{81}}{x^3} + \frac{{10}}{{243}}{x^4} \cr} $$
