If $a,b$, and $c$ are reals satisfying $ \frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=6$, calculate $ x = \frac{(a+b+c)^3}{a^3+b^3+abc}$ $$ \frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=6$$
$$ x = \frac{(a+b+c)^3}{a^3+b^3+abc}$$
As a trivial solution, I found $$a=b=c$$
then, $x$ will always be $9$.
Despite this, my algebra teacher told me there was another solution that I haven't found yet. Is it possible inside the reals? If so, how?
 A: If you put $a=1,b=-2$ and solve for $c$, one root is $c=(17+3\sqrt{33})/2.$ This ends up giving a value for $x$ of about $-101.52$. I think there are lots of ways to make the first equation true, giving various values for $x$.
I gather from Ross Milikan's answer that I misinterpreted the question. I think the OP might mean: if the first symmetric equation in $a,b,c$ holds, and if the value of $x$ turns out to be $9$, find different choices for $a,b,c$. (I had thought it meant "does the relation on $a,b,c$ imply that $x=9$, which it doesn't, as my (simple) example shows.)
So we have $a=b=c=1$ from the OP and Ross' more general case $a=b=c\ne 0$, in which both relations hold, the symmetric relation on $a,b,c$ and the (unsymmetric) relation $x=9.$ This leaves one to wonder: are there any other triples $(a,b,c)$ [not all equal] making both relations hold? It would seem from comments such triples would have some negative entries.
ADDED: I put $c=1$ in both relations sum=6, $x=9$ (OK by homogeneity), and asked maple to solve simultaneously. One answer was the expected $(1,1,1)$, while the other involved a RootOf(**) expression, maple's way of returning the solutions of a polynomial equation. It was a sixth degree polynomial, and when checking the roots numerically it seems all six roots are complex (nonzero imaginary part). So it looks like only the $a=b=c$ solution works for both equations, sum = 6 and $x=9$, even allowing for negative $a,b,c$.
A: You can also have $a=b=c=-2$.  In that case, $x$ is still $9$.  Given the wording of the problem, the value of $x$ should not depend upon what solution you find for $a,b,c$, so you should be able to find one, evaluate $x$ from it, and quit.  If they want yo to prove it independent of $a,b,c$ they should ask.
Added:  $a=b=c$ is a solution for any non-zero $a$, which gives $9$ for $x$  You can probably use the AM-GM inequality to justify that $a=b=c$
