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?

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 $a=b=c$
