Calculate the Taylor series of $f(x) =\ln( 1 -x +x^2) $ and the domain of convergence I just stuck at the following exercise: 
Show that the function f has a Taylor series and calculate it, with $x_0 = 0$.
$$ f(x) = \ln{(1-x+x^2)}$$
Because I already know the Taylor series from 
$$\ln(1+x) = \sum_{n=1}^\infty{(-1)^n \cdot\frac{(x-0)^n}{n}} $$ 
I tried to substitute but surrendered short after the usage of binomial theorem, because I was unable to reorder the sum again.
Also with the 'classic' way, going through the first n derivatives I cannot come to a solution (Wolfram told me one ofc, but I'm unable to reach it)
Any hint(s) and/or solution(s) are welcome :)
 A: A quick solution comes from recognising
$$1-x+x^2 = \frac{1+x^3}{1+x}.$$
With that, we obtain $\ln (1-x+x^2) = \ln (1+x^3) - \ln (1+x)$, and from the Taylor series of $\ln (1+x)$, the Taylor series of $\ln (1+x^3)$ is obtained by replacing $x$ with $x^3$. So we have
\begin{align}
\ln (1-x+x^2) &= \ln (1+x^3) - \ln (1+x)\\
&= \Biggl(\sum_{n=1}^\infty (-1)^{n-1}\frac{x^{3n}}{n}\Biggr) - \Biggl(\sum_{n=1}^\infty (-1)^{n-1}\frac{x^n}{n}\Biggr)\\
&= \sum_{n=1}^\infty a_n x^n
\end{align}
where
$$a_n = \begin{cases} \dfrac{(-1)^n}{n} &, n \not\equiv 0 \pmod{3}\\ \dfrac{(-1)^{n-1}2}{n} &, n \equiv 0 \pmod{3}. \end{cases}$$
Slightly slower, but not reliant on spotting the trick is a factorisation of the argument of the logarithm,
$$1-x+x^2 = (1-\alpha x)\biggl(1-\frac{1}{\alpha}x\biggr),$$
where $\alpha + \frac{1}{\alpha} = 1$, so $\alpha = e^{\pm i\pi/3}$. From that we obtain
$$\ln (1-x+x^2) = -\sum_{n=1}^\infty (\alpha^n + \alpha^{-n}) \frac{x^n}{n}.$$
It is not immediately obvious that these two series are identical, but can be verified without too much difficulty.
