Show that $\frac{1}{x^2+x+1}=\frac{2}{\sqrt3}\sum_{n=0}^{\infty}\sin\bigg(\frac{2\pi(n+1)}{3}\bigg)x^n$ Need to show that:
$$\frac{1}{x^2+x+1}=\frac{2}{\sqrt3}\sum_{n=0}^{\infty}\sin\bigg(\frac{2\pi(n+1)}{3}\bigg)x^n$$
There is a hint given that $x^3-1=(x-1)(x^2+x+1)$ but I don't seem to get how I could use it. If I try to integrate the expression on the left I get:
$$\frac{2}{\sqrt3}\arctan\bigg(\frac{2x+1}{\sqrt3}\bigg)$$
Which has some resemblance to the expression I am trying to come up with.
I know the expansion for $\arctan(x)$ function, but it does not involve sines. So can I arrive at this expression explicitly of should make some useful observations allowing me to incorporate sines into solution?

$$let : C_x = \sum_{n=0}^{\infty} \cos \left( \frac{2\pi(n+1)}{3} \right )x^n$$
$$and : S_x =  \sum_{n=0}^{\infty} \sin \left( \frac{2\pi(n+1)}{3} \right )x^n$$
$$C_x + iS_x = \large{\sum_{n=0}^{\infty} e^{\frac{2\pi i(n+1)}{3}} x^n = e^{i\frac{2\pi}{3}} \sum_{n=0}^{\infty} (xe^{\frac{2\pi i}{3}} )^n}$$
$$= \frac{e^{i\frac{2\pi}{3}}}{1 - xe^{i\frac{2\pi}{3}}} \Rightarrow S_x = \Im \left(   \frac{e^{i\frac{2\pi}{3}}}{1 - xe^{i\frac{2\pi}{3}}} \right ) \overset{\mathbf{do \ it \ by \ yourself}}{=}  = \frac{\sqrt{3}}{2(x^2 + x + 1)}$$
$$\Rightarrow \frac{2}{\sqrt{3}}S_x = \frac{1}{x^2+x+1}$$
$$note : \Im = Im $$
