There are infinitely many possible answers. The simplest approach to $\ln(8/3)$ is to note that $\ln(8/3)=-\ln(3/8)=\ln(1-5/8)$. Now use the ordinary power series for $\ln(1+x)$.
Or else we can use the following classical old trick. We find $x$ such that $$\frac{8}{3}=\frac{1+x}{1-x}.$$ Easily, we get $x=5/11$. Now we use the fact that $$\ln\left(\frac{1+x}{1-x}\right)=\ln(1+x)-\ln(1-x).$$ From the power series expansion of $\ln(1+t)$ we obtain $$\ln(1+x)-\ln(1-x)=2x+\frac{2x^3}{3}+\frac{2x^5}{5}+\frac{2x^7}{7}+\cdots.$$
For something like $\dfrac{1}{\sqrt[3]{2}}$, there are infinitely many possible series. If we really want to be efficient in our calculations, we might note that $$\frac{125}{128}=\frac{125}{64}\frac{1}{2}.$$ Moving things around and taking cube roots we obtain $$\frac{1}{2^{1/3}}=\frac{4}{5}\left(\frac{125}{128}\right)^{1/3}.$$ Now the power series expansion of $(1+t)^{1/3}$, with $t=-3/128$, will get us fast convergence.