The coefficient of $x^2$ in ${(1+x)}^2{(8-x)}^{-\frac{1}{3}}$ is

$\begin{array}{l}{(1+x)}^2{(8-x)}^{-\frac{1}{3}}\\ =(1+2x+x^2)8^{\frac{-1}{3}}{\left(1-\frac{x}{8}\right)}^{\frac{-1}{3}}\\ =\frac{1}{2}(1+2x+x^2)\left(1+\frac{\left(\frac{1}{3}\right)}{1!}.\left(\frac{x}{8}\right)+\frac{\left(\frac{1}{3}\right)\left(\frac{1}{3}+1\right)}{2!}.{\left(\frac{x}{8}\right)}^2+\cdots \right)\\ =\frac{1}{2}\left(1.\frac{1}{288}+\frac{1}{24}.2+1.1\right)\\ =\frac{1}{576}+\frac{1}{24}+\frac{1}{2}=\frac{313}{576}\end{array}$