Answer
$$2.996131$$
Work Step by Step
$$\eqalign{
& \root 3 \of {26.94} \cr
& {\text{Using the result of Taylor polynomial of degree }}4{\text{ at }}x = 0{\text{ found }} \cr
& {\text{in the exercise 20 above }}\left( {page{\text{ 662}}} \right){\text{. for }}f\left( x \right) = \sqrt {x + 27} {\text{ }} \cr
& {\text{we found that}} \cr
& {P_4}\left( x \right) = 3 + \frac{1}{{27}}x - \frac{1}{{2187}}{x^2} + \frac{5}{{531,441}}{x^3} - \frac{{10}}{{43,046,721}}{x^4} \cr
& {\text{To approximate }}\root 3 \of {26.94} ,{\text{ we must evaluate }}\root 3 \of { - 0.06 + 27} ,{\text{ }} \cr
& {\text{then taking }}x = - 0.06 \cr
& {P_4}\left( { - 0.06} \right) = 3 + \frac{1}{{27}}\left( { - 0.06} \right) - \frac{1}{{2187}}{\left( { - 0.06} \right)^2} + \frac{5}{{531,441}}{\left( { - 0.06} \right)^3} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \frac{{10}}{{43,046,721}}{\left( { - 0.06} \right)^4} \cr
& {P_4}\left( {0.03} \right) = 3 - 0.002222222 - 0.00164609 - 0.000203221 \times {10^{ - 6}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 0.003010682 \times {10^{ - 9}} \cr
& {P_4}\left( {0.02} \right) = 2.996131 \cr
& {\text{Thus}}{\text{, }}\root 3 \of {26.94} \approx 2.996131 \cr} $$
