Residue calculations involving logarithms. I am trying to solve a definite integral using residue theory. At some step, I wish to calculate the residue of the function $\frac{\log{(z)}}{z^3+8}$ at $-2e^{2\pi i/3}$. The log function is defined for $0 < \arg z < 2\pi$.
I'm having difficulty doing this simple looking calculation. Can someone please help me with this? 
I have been given a hint that $Res(\frac{\log{(z)}}{z^3+8},z_i)=\frac{-z_i log(z_i)}{24}$. Why and how is this true?

$$\frac{\log(z)}{z^3+8}= \frac{\log(z)}{(z+2e^{\frac{2i\pi}3})(z^2-2e^{\frac{2i\pi}{3}}z+4e^{\frac{-2i\pi}{3}})}$$
Then the residue at $\displaystyle z=-2e^{\frac{2\pi i}{3}}$ should be:
$$\frac{\log(-2e^{\frac{2\pi i}{3}})}{(-2e^{\frac{2\pi i}{3}})^2-2(-2e^{\frac{2\pi i}{3}})(e^{\frac{2i\pi}{3}})+4e^{\frac{-2i\pi}{3}}}= \frac{\log (-2e^{\frac{2i\pi}{3}})}{4e^{\frac{4i\pi}{3}}+ 4e^{\frac{4i\pi}{3}}+ 4e^{\frac{-2i\pi}{3}}}= \frac{\log( -2e^{\frac{2i\pi}{3}})}{8e^{\frac{4i\pi}{3}}+ 4e^{\frac{-2i\pi}{3}}}=  \frac{e^{\frac{2i\pi}{3}}\log( -2e^{\frac{2i\pi}{3}})}{8e^{\frac{6i\pi}{3}}+ 4}=\frac{e^{\frac{2i\pi}{3}}\log( -2e^{\frac{2i\pi}{3}})}{12}$$
