<s>[INST] Find the power series of $f(x)=\frac{1}{x^2+x+1}$ I want to find the power series of $$f(x)=\frac{1}{x^2+x+1}$$
How can I prove the following?
$$f(x)=\frac{2}{\sqrt{3}} \sum_{n=0}^{\infty} \mathrm{sin}\frac{2\pi(n+1)}{3} x^n  \,\,\,\, |x|<1$$
In particular I would like to know how to proceed in this case. The polinomial $x^2+x+1$ has no roots so here I cannot use partial fraction decomposition: what method should I use?
 [/INST] The polynomial $x^2+x+1=\Phi_3(x)$ has no real roots, but it vanishes at $x=e^{\pm\frac{2\pi i}{3}}$. In particular, by setting $\omega=\exp\left(\frac{2\pi i}{3}\right)$ and $\overline{\omega}=\omega^2=\exp\left(\frac{4\pi i}{3}\right)$,
$$ \frac{1}{x^2+x+1} = \frac{1}{(x-\omega)(x-\omega^2)} = \frac{i\omega^2}{\sqrt{3}}\cdot\frac{1}{1-\omega^2 x}-\frac{i\omega}{\sqrt{3}}\cdot\frac{1}{1-\omega x} $$
where the RHS, expanded as the difference between two geometric series, equals
$$ \frac{i}{\sqrt{3}}\sum_{n\geq 0}\left(\omega^{2n+2}-\omega^{n+1}\right)x^n =\frac{2}{\sqrt{3}}\sum_{n\geq 0}\sin\left(\frac{2\pi(n+1)}{3}\right)x^n$$
as wanted. That clearly simplifies, since
$$ \frac{1}{1+x+x^2}=\frac{1-x}{1-x^3} = \sum_{m\geq 0}\left(x^{3m}-x^{3m+1}\right),$$
too, and the Taylor series at the origin is unique.
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