Answer
$$ \approx 0.94592$$
Work Step by Step
$$\eqalign{
& {0.92^{2/3}} \cr
& {\text{Using the result of Taylor polynomial of degree }}4{\text{ at }}x = 0{\text{ found }} \cr
& {\text{in the exercise 23 above }}\left( {page{\text{ 662}}} \right){\text{. for }}f\left( x \right) = {\left( {1 + x} \right)^{2/3}} \cr
& {\text{we found that}} \cr
& {P_4}\left( x \right) = 1 + \frac{2}{3}x - \frac{1}{9}{x^2} + \frac{4}{{81}}{x^3} - \frac{7}{{243}}{x^4} \cr
& {\text{To approximate }}{0.92^{2/3}},{\text{ we must evaluate }}{\left( {1 + \left( { - 0.08} \right)} \right)^{2/3}},{\text{ }} \cr
& {\text{then taking }}x = - 0.08 \cr
& {P_4}\left( { - 0.08} \right) = 1 + \frac{2}{3}\left( { - 0.08} \right) - \frac{1}{9}{\left( { - 0.08} \right)^2} + \frac{4}{{81}}{\left( { - 0.08} \right)^3} - \frac{7}{{243}}{\left( { - 0.08} \right)^4} \cr
& {P_4}\left( { - 0.08} \right) = 0.94592 \cr
& {\text{Thus}}{\text{, }}{0.92^{2/3}} \approx 0.94592 \cr} $$
