How to show that a rational polynomial is irreducible in $\mathbb{Q}[a,b,c]$? For example, I try to show this polynomial $$p(a,b,c)=a(a+c)(a+b)+b(b+c)(b+a)+c(c+a)(c+b)-4(a+b)(a+c)(b+c)(*)$$ is irreducible, where $a,b,c\in \mathbb{Q}$.
The related problem is Ask for the rational roots of $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=4.$. Could I consider the points $(*)$ intersect with $L_{\infty}$ are three? $L_{\infty}$ is the infinity line in a projective space $\mathbb{C}P^2$.
On
To ask about complete (three linear factor) reducibility over the complexes, we take the Hessian matrix of second partials. The entries are linear in the named variables. Next, let $\Delta$ be the determinant of the Hessian. This $\Delta$ is once again a cubic form. The original cubic (homogeneous) ternary form factors completely if and only if $\Delta $ is a constant multiple of it. I have finished this first test, your cubic does not factor completely.
More difficult if the cubic might be a linear times an irreducible quadratic. In that case, there is still a conclusive test:
For your problem, the coefficients in order from $0$ to $9$ are $$ 1,-3,-3,-3,-5,-3,1,-3,-3,1 $$ where the best looking diagram is a triangle as in bowling. The $-5$ refers to $-5abc \; ,$ the $1$s refer to $a^3,b^3,c^3 \; .$
Alright, I wote a little program to correctly type the 8 by ten matrix, then put that into gp-Pari. The original polynomial is irreducible over the complex numbers.
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? sch = [ 0, 3, 0, -6, -6, 0, -3, -5, -3, 0; 0, 0, 3, 0, -6, -6, 0, -3, -5, -3; -3, -6, -5, 3, -6, -3, 0, 0, 0, 0; 0, 0, -3, 0, -6, -5, 0, 3, -6, -3; -3, -5, -6, -3, -6, 3, 0, 0, 0, 0; 0, -3, 0, -5, -6, 0, -3, -6, 3, 0; 1, 0, -3, 3, 0, -3, -2, 3, 0, 1; 1, -3, 0, -3, 0, 3, 1, 0, 3, -2]
%25 =
[ 0 3 0 -6 -6 0 -3 -5 -3 0]
[ 0 0 3 0 -6 -6 0 -3 -5 -3]
[-3 -6 -5 3 -6 -3 0 0 0 0]
[ 0 0 -3 0 -6 -5 0 3 -6 -3]
[-3 -5 -6 -3 -6 3 0 0 0 0]
[ 0 -3 0 -5 -6 0 -3 -6 3 0]
[ 1 0 -3 3 0 -3 -2 3 0 1]
[ 1 -3 0 -3 0 3 1 0 3 -2]
? matrank(sch)
%26 = 8
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Suppose contrary that $p(a,b,c)$ is reducible over $\mathbb{Q}$. You can write $p(a,b,c)$ as $$a^3+b^3+c^3-3(b+c)a^2-3(c+a)b^2-3(a+b)c^2-5abc\,.$$ It suffices to regard $p(a,b,c)$ as a polynomial over $\mathbb{F}_3$ (why?). Over $\mathbb{F}_3$, $$p(a,b,c)=a^3+b^3+c^3+abc=a^3+(bc)a+(b+c)^3\,.$$ Since $p(a,b,c)$ is homogeneous of degree $3$ and reducible, it has a linear factor $a+ub+vc$ for some $u,v\in\mathbb{F}_3$. Clearly, we must have $ub+vc \mid (b+c)^3$, whence $u=v=1$ or $u=v=-1$. However, both choices are impossible via direct computation.