Answer
$2+\frac{x}{12}+2\Sigma_{n=2}^{\infty}(-1)^{n+1}\frac{2.5.8.......(3n-4)x^{n}}{24^{n}n!}$
and
$R=8$
Work Step by Step
$\sqrt[3] {8+x}=(8+x)^{1/3}=(8+x)^{1/3}=2(1+\frac{x}{8})^{1/3}$
$=2+\frac{x}{12}+2\Sigma_{n=2}^{\infty}(-1)^{n+1}\frac{2.5.8.......(3n-4)x^{n}}{24^{n}n!}$
$$\lim\limits_{n \to \infty}|\dfrac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{\frac{2.5.8.......(3n-4)(3(n+1)-4)x^{n+1}}{24^{n+1}(n+1)!}}{\frac{2.5.8.......(3n-4)x^{n}}{24^{n}n!}}|$$
$=\lim\limits_{n \to\infty}|\frac{(3n-1)x}{24(n+1)}|$
$=|\frac{x}{8}|$
The series will converge when $|\frac{x}{8}|\lt 1$, or $|x|\lt 8$ so $R=8$.
