Show that $r^{2} = 3$ from a geometric series 
The sum of the first two terms of geometric sequence $a_n=a_0r^n$ is $7$, and the sum of the first six terms is $91$.
Show that $r^{2} = 3$.

How do we go about this?
Here's what I have tried -
$\displaystyle \Rightarrow \frac{\dfrac{a(r^{6}-1)}{r-1}}{\dfrac{a(r^{2}-1)}{r-1}} = \frac{91}{7}$
$\displaystyle \Rightarrow \frac{(r^{6}-1)}{(r^{2}-1)} = \frac{91}{7}$
$\Rightarrow\;???$
Any help would be highly appreciated.

Use this identity $a^3- b^3 = (a-b)(a^2 +ab+b^2)$, and then solve a quadratic equation.
$$\Rightarrow \frac{(r^{6}-1)}{(r^{2}-1)} = 13$$
$$\Rightarrow \frac{(r^{2}-1)(r^4+r^2+1)}{(r^{2}-1)} = 13$$
$$\Rightarrow r^4+r^2+ 1= 13$$
$$\Rightarrow r^4+r^2-12=0$$
You'll find something like $(r^2 -3)(r^2 +4)=0$
BINGO!
