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})] $

Question

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 $$ \left(\frac {1}{1-r_{+}x}\right) - \left(\frac {1}{1-r_{-}x}\right)= {\sum_{n\ge0}\ r_+^nx^n}-{\sum_{n\ge0}\ r_-^nx^n} $$

Answer 1

\begin{align*} \frac {1}{\sqrt{5}}\left(\frac {1}{x+r_+} - \frac {1}{x+r_-}\right) &= \frac {1}{\sqrt{5}} \frac{x}{x} \left(\frac {1}{x+r_+} - \frac {1}{x+r_-}\right) &&,x \neq 0 \\ &= \frac {1}{\sqrt{5}\, x} \left(\frac {x}{x+r_+} - \frac {x}{x+r_-}\right) &&,x \neq 0 \\ &= \frac {1}{\sqrt{5}\, x} \left(\frac {(x^2 + r_- x) - (x^2 + r_+ x)}{(x+r_+)(x+r_-)}\right) &&,x \neq 0 \\ &= \frac {1}{\sqrt{5}\, x} \left(\frac {r_- x - r_+ x}{(x+r_+)(x+r_-)}\right) &&,x \neq 0 \\ &= \frac {1}{\sqrt{5}\, x} \left(\frac {r_- x - r_+ x}{x^2 + x - 1}\right) &&,x \neq 0 \\ &= \frac {1}{\sqrt{5}\, x} \left(\frac {r_- x - r_+ x}{-(-x^2 - x + 1)}\right) &&,x \neq 0 \\ &= \frac {1}{\sqrt{5}\, x} \left(\frac {-r_- x + r_+ x}{-x^2 - x + 1}\right) &&,x \neq 0 \\ &= \frac {1}{\sqrt{5}\, x} \left(\frac {1-r_- x -1 + r_+ x}{(1-r_+ x)(1-r_- x)}\right) &&,x \neq 0 \\ &= \frac {1}{\sqrt{5}\, x} \left(\frac {(1-r_- x) - (1 - r_+ x)}{(1-r_+ x)(1-r_- x)}\right) &&,x \neq 0 \\ &= \frac {1}{\sqrt{5}\, x} \left(\frac {1}{1-r_+ x} - \frac {1}{1-r_- x}\right) &&,x \neq 0 \\ \end{align*}

For completeness, we verify that the RHS of the equation sought is undefined for $x=0$, so the condition cannot be removed.

Answer 2

From your last step:

$$\frac{1}{\sqrt{5}}\left(\frac{r_{+}}{1 - r_{+}x} - \frac{r_{-}}{1 - r_{-}x}\right)$$ $$= \frac{1}{\sqrt{5}x}\left(\frac{r_{+}x}{1 - r_{+}x} - \frac{r_{-}x}{1 - r_{-}x}\right)$$ $$= \frac{1}{\sqrt{5}x}\left(\left[\frac{1}{1 - r_{+}x} - 1\right] - \left[\frac{1}{1 - r_{-}x} - 1\right]\right)$$ $$= \frac{1}{\sqrt{5}x}\left(\frac{1}{1 - r_{+}x} - \frac{1}{1 - r_{-}x}\right)$$