Answer
$L\left( x \right) = \frac{1}{{12}}x + \frac{4}{3}$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \root 3 \of x ,{\text{ }}a = 8 \cr
& {\text{Evaluate }}f\left( x \right){\text{ at }}a = 8 \cr
& f\left( 8 \right) = \root 3 \of 8 \cr
& f\left( 8 \right) = 2 \cr
& {\text{Differentiating the given function}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\root 3 \of x } \right] \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^{1/3}}} \right] \cr
& f'\left( x \right) = \frac{1}{3}{x^{1/3 - 1}} \cr
& f'\left( x \right) = \frac{1}{3}{x^{ - 2/3}} \cr
& f'\left( x \right) = \frac{1}{{3{x^{2/3}}}} \cr
& {\text{Evaluate }}f'\left( x \right){\text{ at }}a = 8 \cr
& f'\left( 8 \right) = \frac{1}{{3{{\left( 8 \right)}^{2/3}}}} \cr
& f'\left( 8 \right) = \frac{1}{{12}} \cr
& {\text{The linear approximation at }}f\left( a \right){\text{ is given by: }} \cr
& L\left( x \right) = f\left( a \right) + f'\left( a \right)\left( {x - a} \right){\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Substituting }}f\left( 8 \right) = 2,{\text{ }}f'\left( 8 \right) = \frac{1}{{12}}{\text{ and }}a = 8{\text{ into }}\left( {\bf{1}} \right) \cr
& L\left( x \right) = 2 + \frac{1}{{12}}\left( {x - 8} \right) \cr
& {\text{Simplifying}} \cr
& L\left( x \right) = 2 + \frac{1}{{12}}x - \frac{2}{3} \cr
& L\left( x \right) = \frac{1}{{12}}x + \frac{4}{3} \cr} $$