Show that $\frac {1}{\sqrt{5}}[(\frac {1}{x+r_+}) - (\frac {1}{x+r_-}) = \frac {1}{\sqrt{5}x}[(\frac {1}{1-r_{+}x}) - (\frac {1}{1-r_{-}x})] $ I need to manipulate this equation:
$$
\frac {1}{\sqrt{5}}\left(\frac {1}{x+r_+} - \frac {1}{x+r_-}\right) 
$$
to show that
$$ \frac {1}{\sqrt{5}}\left(\frac {1}{x+r_+} - \frac {1}{x+r_-}\right) = \frac {1}{\sqrt{5}x}\left(\frac {1}{1-r_{+}x} -\frac {1}{1-r_{-}x}\right)
$$ (*)
where$$r_+=\frac {1+\sqrt{5}}{2} $$and $$r_-=\frac {1-\sqrt{5}}{2}$$ 
I know that both RHS and LHS of (*) are equal to $\frac{1}{1-x-x^2}  $ , but don't know how we get RHS manipulating LHS before someone told us that RHS is also equal to $\frac{1}{1-x-x^2} $ 
Edit:
Since $$r_+r_-=-1$$, I replace $r_+$ with $r_-$ and $r_-$ with $ r_+$ in the LHS of (*)
So that, I get
$$ \frac {1}{\sqrt{5}}\left(\frac {r_+}{1-r_+x} - \frac {r_-}{1-r_-x}\right)$$ Still, I am not there. Can anybody help me at this step?
After showing the equality above I can continue as
$$
\left(\frac {1}{1-r_{+}x}\right) - \left(\frac {1}{1-r_{-}x}\right)= {\sum_{n\ge0}\ r_+^nx^n}-{\sum_{n\ge0}\ r_-^nx^n}
$$

From your last step:
$$\frac{1}{\sqrt{5}}\left(\frac{r_{+}}{1 - r_{+}x} - \frac{r_{-}}{1 - r_{-}x}\right)$$ 
$$= \frac{1}{\sqrt{5}x}\left(\frac{r_{+}x}{1 - r_{+}x} - \frac{r_{-}x}{1 - r_{-}x}\right)$$
$$= \frac{1}{\sqrt{5}x}\left(\left[\frac{1}{1 - r_{+}x} - 1\right] - \left[\frac{1}{1 - r_{-}x} - 1\right]\right)$$
$$= \frac{1}{\sqrt{5}x}\left(\frac{1}{1 - r_{+}x} - \frac{1}{1 - r_{-}x}\right)$$
