Algebraic Solving Contest Problem The problem is as follows
If $x^2+x-1=0$, compute all possible values of $\frac{x^2}{x^4-1}$
This was a no-calculator 10 min for 2 problem format contest.
I started by using quadratic formula, but the answers I got were too ugly to be plugged into the equation before time ran out. I tried several algebraic manipulations, but after losing time on the other problem, i found no quick answer to this. I presume there is one step i'm missing to make that leap. Can this be done easily?
 A: $$x^2=1-x$$
$$\implies\frac{x^2}{x^4-1}=\frac{1-x}{(1-x)^2-1}=\frac{1-x}{x^2-2x}=\frac{1-x}{1-x-2x}$$
$$=\frac{1-x}{1-3x}\ \ \ \ (1)$$
$$=\frac13\left(1+\frac2{1-3x}\right)\  \ \ \ (2)$$
Now put the two values of $x$, one by one in either $(1)$ or in $(2)$ which ever you like
A: Use the quadratic formula and solve for the zeros. 
Verify that they are $x = \frac{1}{2}(-1-\sqrt{5}),\frac{1}{2}(\sqrt{5}-1)$.
Now just evaluate the later for these values of $x$.
$\textbf{Extra}$: You can also complete the square and you will get: $(x+\frac{1}{2})^2 -\frac{5}{4}=0$, then solve for $x$.
A: If $x^2+x-1=0$, then $x^2=1 – x$
Squaring both sides, we get $x^4=1 – 2x + x^2$
∴ $x^4 – 1 = x^2 – 2x$
$= (1 – x) – 2x$
$= 1 – 3x$
Thus, the expression $=$ ${1 – x} \over {1 – 3x}$
By long division, it is equivalent to ${1} \over {3}$+ ${2}\over {3(1 – 3x)}$
Edit (1) - Please ignore the following:-
As pointed out, $x^4 – 1$ can also be factorized as $(x^2+1)(x+1)(x-1)$.
Thus, the expression can take any value of x except 1, -1, or 1/3 (if real values of x are assumed).
Edit (2) - Please add the following:-
As pointed out, $x = \frac{1}{2}(-1 \pm \sqrt{5})$
Just put $x = \frac{1}{2}(-1 + \sqrt{5})$ in the simplified expression and do a little rationalization first. For the $x = \frac{1}{2}(-1 - \sqrt{5})$, the result is very similar.
Using the simplified expression, 10 minutes should be sufficient. 
A: Let $a, b$ be the roots of $x^2+x-1=0$.  Then $a+b = -1, ab = -1$.  For simplicity,  consider now the reciprocals of the values we wish to find, i.e. $\dfrac{a^4-1}{a^2} = k_1$ and $\dfrac{b^4-1}{b^2} = k_2$, then 
$$k_1 + k_2 = a^2+b^2-\left(\frac1{a^2}+\frac1{b^2} \right), \quad k_1 k_2 = \frac{(a^4-1)(b^4-1)}{(ab)^2}$$
Now $a^2+b^2 = (a+b)^2-2ab = 3$, so $k_1 + k_2 = 3-3=0$.
Also $(a^4-1)(b^4-1) = 1+(ab)^4-(a^4+b^4) = 1+1-(3^2-2) = -5$, so $k_1 k_2 = -5$.
So $k_1, k_2$ are roots of $y^2-5=0 \implies \{k_1, k_2\} = \{\sqrt5, -\sqrt5\}$.  Take reciprocals now.
