How does this proof of Fibonacci work

Question

\begin{eqnarray*} F_{i+1}&=&F_{i} + F_{i-1}\\ &=&\frac{\phi^i-\hat{\phi^i}}{\sqrt5}+\frac{\phi^{i-1}-\hat{\phi^{i-1}}}{\sqrt5}\\ &=&\frac{\left(\phi+\hat{\phi}\right)\left(\phi^i-\hat{\phi^i}\right)-\phi\hat{\phi}\left(\phi^{i-1}-\hat{\phi^{i-1}}\right)}{\sqrt5}\text{(Why does this work?)}\\ &=&\frac{\phi^{i+1}-\phi\hat{\phi^i}+\hat{\phi}\phi^i-\hat{\phi^{i+1}}-\phi^i\hat{\phi}+\phi\hat{\phi^i}}{\sqrt5}\\ &=&\frac{\phi^{i+1}-\hat{\phi^{i+1}}-\phi\hat{\phi^i}+\phi\hat{\phi^i}-\hat{\phi}\phi^i+\hat{\phi}\phi^i}{\sqrt5}\\ &=&\frac{\phi^{i+1}-\hat{\phi^{i+1}}}{\sqrt5}\\ \end{eqnarray*}

How to does the following become simplified?

$\frac{\phi^i-\hat{\phi^i}}{\sqrt5} =\left(\phi+\hat{\phi}\right)\left(\phi^i-\hat{\phi^i}\right)$

$\frac{\phi^{i-1}-\hat{\phi^{i-1}}}{\sqrt5} = \phi\hat{\phi}\left(\phi^{i-1}-\hat{\phi^{i-1}}\right)$

Answer 1

$\phi$ and $\hat\phi$ are the two roots of $x^2 - x - 1 = 0$. Explicitly, $$\phi = \frac{1 + \sqrt{5}}{2},\ \ \ \ \ \hat\phi = \frac{1 - \sqrt{5}}{2}$$ Then it follows that $$\phi + \hat\phi = 1,\ \ \ \ \phi\hat\phi = \frac{1-5}{4} = -1$$

Answer 2

HINT:

$$\hat{\phi} = - \frac{1}{\phi} \quad \quad \text{and} \quad \quad \hat{\phi} + \phi = 1$$

So multiplying by 1, won't change anything.

Above identities can be easily proven when you substitute:

$$\phi = \frac{1 + \sqrt{5}}{2} \quad \text{and} \quad \hat{\phi} = \frac{1 - \sqrt{5}}{2}$$