Generating Functions- Closed form of a sequence We are given the following generating function : $$G(x)=\frac{x}{1+x+x^2}$$
The question is to provide a closed formula for the sequence it determines.
I have no idea where to start. The denominator cannot be factored out as a product of two monomials with real coefficients. Any sort of help to solve this problem is welcome!

If you want the Binet formula for this you can solve the matrix
Set up the partial fraction 
$$\frac{x}{1+x+x^2}=\frac{A}{1-ax}+\frac{B}{1-bx}$$ where $$a=\frac{-1+\sqrt{3}i}{2}$$ and $$ b=\frac{-1-\sqrt{3}i}{2}$$ so that
$$A(1-bx)+B(1-ax)$$
The matrix becomes
\begin{bmatrix}
-b&-a&1\\[0.3em]
1&1&0
 \end{bmatrix}
or
\begin{bmatrix}
\frac{1+\sqrt{3}i}{2}&\frac{1-\sqrt{3}i}{2}&1\\[0.3em]
1&1&0
 \end{bmatrix}
which in reduced echelon form is 
\begin{bmatrix}
1&0&\frac{-\sqrt{3}i}{3}  \\[0.3em]
0&1&\frac{\sqrt{3}i}{3}
\end{bmatrix}
The closed form then becomes
$$a_n=\frac{-\sqrt{3}i}{3}\left(\frac{-1+\sqrt{3}i}{2}\right)^n+\frac{\sqrt{3}i}{3}\left(\frac{-1-\sqrt{3}i}{2}\right)^n$$
$a_0=0$ $a_1=1$, $a_2=-1$, $a_3=0$, $a_4=1$, $a_5=-1$ etc.
