I always liked programming, even though I was far from an expert. So this was my plan B, but never had to actually implement it. Teaching at a liberal arts college is a good option for someone who likes teaching and is good at it. Wall Street is an option for some. It really depends on what your strengths and interests are.

@joro You mean apart from the silly stuff like $a=kc,b=kd$?

There is some value in checking that a given argument extends to a slightly more general setting. You can not expect it to be published in a particularly good journal, but your first paragraph makes it sound like the research is legitimate. You can always show it around to experts in the field to see if they view it "too obvious to publish".

I am afraid not.

There is definitely more than one way of writing such expressions, even for $k=2$. Do you want an expression with some specific properties?

At the very least, you would have to have a version of Euler's theorem for all $a$ rather than the ones coprime to $n$. If you can formulate it, then it is conceivable that some inductive argument would also work.

Your conjecture.

Where does this condition come from? It looks decidedly unusual to me.

Correct. The discriminant is $(x-y)^2(y-z)^2(x-z)^2$, so if the roots are real, then it is nonnegative.

Have you tried solving it by brute force calculation? You can parameterize the conic, pick $A_1,A_2,A_3,A_4,A_5,A_6,B_1$ for generic values of parameters, and proceed to calculate $B_2,...,B_6$, as well as $O_1,...,O_6$. It might be beyond the usual software's capabilities, but it might not be.

The most elegant solution is to write "This is an easy exercise which we leave to the reader."

It is easy to see by convexity of sorts that there are no other solutions near $(1,1,1)$. Specifically, let $x=e^a$, $y=e^b$, $z=e^c$. Then you have to solve $e^{2a}-3e^{a} + .... = -6$ for $a+b+c=0$. By Jensen's you will not have nontrivial solutions in the region where $f(t)=e^{2t}-3e^t$ is convex.

You can do Lagrange multiplers for $xyz$ on the sphere. This will give you three equations like $xy=(2z-3)\lambda$ and so on. By subtracting one of such equation from the other you get $x=z$ or $y=-2\lambda$. Either way you get linear relations on $x,y,z$ and it should be easy to solve. Then you will get $xyz\geq 1$ with minimum at $(1,1,1)$.

You can try to take log coords $x=e^a,y=e^b,z=e^c$ and then see if the inequality $(e^a-3/2)^2+(e^b-3/2)^2+(e^c-3/2)^2\leq 3/4$ gives a convex shape. But I think that my approach with the discriminant is short, if not "nice".