<s>[INST] 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.
 [/INST] Note that $$r^6-1=(r^3+1)(r^3-1).$$
From @MMM's answer, we can observe that for all $a$ and $b$, $$a^3-b^3=(a-b)(a^2+ab+b^2). $$ Let $a=r$ and $b=\pm 1$. $$\therefore r^6-1=(r+1)(r^2-r+1)\cdot (r-1)(r^2+r+1).$$ Since $r^2-1=(r+1)(r-1)$, it follows then that $$(r^2-r+1)(r^2+r+1)=13.$$
Can you take it from here? :)
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