How to show a rational polynomial is irreducible in $\mathbb{Q}[a,b,c]$?
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.
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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