Factoring polynomials to solve for variables? I have two equations: 
$3192 =\frac{a\left(r^6-1\right)}{r-1}$
$3704 = \frac{a\left(r^7-1\right)}{r-1}$
I need to solve for $r$ and $a$. How would I do so? Substitution leads to a really messy equation.
 A: By manipulation we get $3704(r^6-1)=3192(r^7-1)$, except that $r=1$ should presumably not be viewed as a solution of the original equation.
So in principle we can divide both sides by $r-1$ and reach an equation of degree $6$. 
But let's not do this, just rewrite as $3192r^7 -3704r^6+512=0$, or more simply 
$399r^7-463r^6 +64=0$.  Emboldened by the simple nature of $64$, we may look for a rational root other than $1$. A bit of fooling around yields the root $r=-2/3$. Perhaps that root is useful to you, though probably not, since it is negative.  It is easy to find the corresponding $a$.
Now (in principle) divide the polynomial $399r^7-463r^6 +64$ by $r-1$, then by $r+2/3$. We reach a degree $5$ polynomial that does not appear to have a special enough structure.  I did not find another rational root. So we end up with a quintic. There is another real root, which can be approximated by a numerical procedure, but which I do not have a closed form for.
Remark: I did a rough calculation. The positive root is fairly close to $1$, a bit under.  Dividing by $r-1$ is useful so that a numerical method will not get wrongly sucked into $1$. 
A: The 'trick' here is that $\frac{r^6-1}{r-1}$ and $\frac{r^7-1}{r-1}$ are are both partial sums of geometric series; $\frac{r^7-1}{r-1}=r^6+r^5+r^4+r^3+r^2+r+1$ and similarly for the other term, so by taking the difference you get $ar^6=512$; plugging this into either equation gives a polynomial you can solve to find the relevant values of $r$.  From there, you can follow along with Andre's answer.
