Use the binomial series to expand the function as a Maclaurin series and to find the second Taylor polynomial.

$\frac{1}{\sqrt[3]{8+x}}$

$T_2=-\frac{4}{288}{x}^2+\frac{1}{6}x-1$

$T_2=-\frac{4}{2,304}{x}^2+\frac{1}{48}x-\frac{1}{2}$

$T_2=\frac{4}{2,304}{x}^2-\frac{1}{48}x+\frac{1}{2}$

$T_2=\frac{4}{288}{x}^2-\frac{1}{6}x+1$