Maclaurin Series Expansion $\ln(8+x^3)$. 
Find Maclaurin Series Expansion of function $\ln(8+x^3)$. For which $x$ series converges? For which $ x $ ,$ f(x)$ equals sum of this series?


$$ (\ln(8+x^3))'=\frac{3x^2}{8+x^3} =\frac{\frac{3}{8}x^2}{1-(-\frac{x^3}{8})}=$$
$$ =\sum_{n=1}^{\infty}\frac{3x^2}{8}(-1)^n\frac{x^{3n}}{8^n}$$
and 
$$ \sum_{n=1}^{\infty}\int\frac{3x^2}{8}(-1)^n\frac{x^{3n}}{8^n}dx=$$
$$=\sum_{n=1}^{\infty}\frac{(-1)^n3}{8^{n+1}}\int x^{3n+2}dx=$$
$$=\sum_{n=1}^{\infty}\frac{(-1)^n3}{8^{n+1}}\frac{x^{3n+3}}{3n+3}+C$$
for $x=0$
$$\ln(8+0^3)= \sum_{n=1}^{\infty}\frac{(-1)^n3}{8^{n+1}}\frac{0^{3n+3}}{3n+3}+C $$
$$ C=0 $$
$8+x^3>0 $
$x>-2 $
and
$|\frac{-x^3}{8}|<1$
$x<2$

I have to find if it converges only at $x=2$, $x=-2$ is out of domain?  For which $x$ series converges? For which $ x $ ,$ f(x)$ equals sum of this series? From Abel theorem $\forall$ x which converges $f(x)$=sum of series?


HINT:
$$\ln(8+x^3)=\ln8+\ln\left(1+\dfrac{x^3}8\right)$$
See Why does the taylor series of $\ln (1 + x)$ only approximate it for $-1<x \le 1$?
OR
Taylor series for $\log(1+x)$ and its convergence
OR
What is the correct radius of convergence for $\ln(1+x)$?
