Answer
$2,~\frac{1}{12}(x-8),~-\frac{1}{288}(x-8)^2,~\frac{5}{20736}(x-8)^3$
Work Step by Step
Let's suppose $n=0,1,2,3$
then,
$f^n(x)=f^n(8)$
$f(x)=\sqrt[3] x$
$=\sqrt[3] 8$
$=(8)^\frac{1}{3}$
$=(2^3)^\frac{1}{3}$
$f(x)=2$
$f'(x)=\frac{d}{dx}\sqrt[3] x$
$~~~~~~~~~=\frac{1}{3}(x)^{\frac{1}{3}-1}$
$~~~~~~~~~=\frac{1}{3}(x)^{\frac{1-3}{3}}$
$~~~~~~~~~=\frac{1}{3}(x)^{\frac{-2}{3}}$
$f'(x)=\frac{1}{3(x)^{\frac{2}{3}}}$
$f'(8)=\frac{1}{3(8)^{\frac{2}{3}}}$
$f'(8)=\frac{1}{12}$
$f''(x)=\frac{d}{dx}(\frac{1}{3(x)^{\frac{2}{3}}})$
$~~~~=\frac{1}{3}\frac{d}{dx}(x)^{\frac{-2}{3}}$
$~~~~=\frac{-2}{9}(x)^{\frac{-2}{3}-1}$
$~~~~=\frac{-2}{9}(x)^{\frac{-2-3}{3}}$
$~~~~=\frac{-2}{9}(x)^{\frac{-5}{3}}$
$f''(x)~=-\frac{2}{9(x)^{\frac{5}{3}}}$
$f''(8)=-\frac{2}{9(8)^{\frac{5}{3}}}$
$f''(8)=-\frac{2}{288}$
$f'''(x)=\frac{d}{dx}(-\frac{2}{9(x)^{\frac{5}{3}}})$
$~~~~~~~~~~~=\frac{10}{27}(x)^{\frac{-5}{3}-1}$
$~~~~~~~~~~~=\frac{10}{27}(x)^{\frac{-5-3}{3}}$
$~~~~~~~~~~~=\frac{10}{27}(x)^{\frac{-8}{3}}$
$f'''(x)=\frac{10}{27(x)^{\frac{8}{3}}}$
$f'''(8)=\frac{10}{27(8)^{\frac{8}{3}}}$
$f'''(8)=\frac{10}{6912}$
The $nth$ term of Taylor Series $=\frac{f^{n-1}(8)x^{n-1}}{(n-1)!}$
The $1st$ term of Taylor Series $=\frac{f(8)x^{0}}{0!}$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\frac{2(1)}{0!}$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\frac{2}{0!}$
The $1st$ term of Taylor Series $=2$
The $2nd$ term of Taylor Series $=\frac{f(8)x^{1}}{1!}$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\frac{\frac{1}{12}}{1!}(x-8)^1$
The $2nd$ term of Taylor Series $=\frac{1}{12}(x-8)$
The $3rd$ term of Taylor Series $=\frac{f(8)x^{2}}{2!}$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\frac{-\frac{2}{288}}{2!}(x-8)^2$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=-\frac{1}{288}(x-8)^2$
The $3rd$ term of Taylor Series $=-\frac{1}{288}(x-8)^2$
The $4th$ term of Taylor Series $=\frac{f(8)x^{3}}{3!}$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\frac{\frac{10}{6912}}{3!}(x-8)^3$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\frac{5}{20736}(x-8)^3$
The $4th$ term of Taylor Series $=\frac{5}{20736}(x-8)^3$
Hence, the $4th$ non zero Taylor Series is :
$2,~\frac{1}{12}(x-8),~-\frac{1}{288}(x-8)^2,~\frac{5}{20736}(x-8)^3$