Answer
$\dfrac{1}{3}\left[\ln (x-2)+\ln {(x+1})-2\ln (x+4)\right] $
Work Step by Step
We use the rules of logarithms to obtain:
$\ln \left[\dfrac {x^2-x-2}{(x+4)^2}\right]^{\frac{1}{3}}\\=\dfrac{1}{3}\ln \left[\dfrac {x^2-x-2}{(x+4)^2}\right]\\=\dfrac{1}{3}\left[ \ln (x-2)(x+1)-2\ln (x+4) \right]\\=\dfrac{1}{3}\left[\ln (x-2)+\ln {(x+1})-2\ln (x+4)\right] $
