Proving that a polynomial about the volume of a tetrahedron is irreducible We know that the volume of a tetrahedron $ABCD$ can be represented as
$$144V^2=(a^2b^2d^2+b^2c^2e^2+c^2a^2f^2+b^2a^2e^2+c^2b^2f^2+a^2c^2d^2+c^2e^2f^2+a^2f^2d^2+b^2d^2e^2+c^2d^2f^2+a^2e^2d^2+b^2f^2e^2)-(a^2b^2c^2+a^2e^2f^2+b^2f^2d^2+c^2d^2e^2)$$
where $a=BC, b=CA, c=AB, d=AD, e=BD, f=CD$.
Let $F(a,b,c,d,e,f)$ be the right hand side of the above equation. Then, here is my question.

Question : Is $F(a,b,c,d,e,f)$ irreducible as a polynomial?

The answer must be YES, but I don't know how to prove that. Can anyone help?

Notice there are no linear terms in $a$. So it is a polynomial of the form $Aa^2+B$. So, $B/A$ must be a square.
$A=b^2(d^2+e^2+f^2e^2+c^2)+(e^2f^2+c^2f^2+c^2d^2+f^2d^2+c^2d^2f^2+e^2d^2)$
and 
$B=b^2(f^2d^2+c^2e^2+c^2f^2+d^2e^2)+(c^2d^2e^2+c^2e^2f^2)$
if I didn't miss some term.
Check that $A$ and $B$ don't have a common factor, which I haven't done. And then it is clear they are not squares themselves, since they have no terms linear in $b$.
