It is easy to see that $$a^2+b^2+c^2+ab+bc+ca=\frac{1}{2}((a+b)^2+(b+c)^2+(c+a)^2),$$ $$a^2+b^2+c^2+2(ab+bc+ca)=(a+b+c)^2,$$ $$a^2+b^2+c^2-ab-bc-ca=\frac{1}{2}((a-b)^2+(b-c)^2+(c-a)^2),$$ and moreover, these are all of the form $a^2+b^2+c^2+m(ab+bc+ca)$, where $m=1,2,-1$. So what "factorizations" are there for $a^3+b^3+c^3+mabc$? We know that for $m=-3$ we get $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca).$$ Moreover, if we call $a^2+b^2+c^2+m(ab+bc+ca)=A_m$ and $a^3+b^3+c^3+mabc=B_m$, it is interesting to note that $B_{-3}=(a+b+c)A_{-1}$. But this isn't my question. I am trying to find similar factorizations for $B_m$ to those of $A_m$. (I do not know of a better way to describe what I mean by "factorization", but I am hoping the examples make it clear.)
Just post anything you come up with, even if $m$ is not integral, etc.
Also, what would be the analog(s) of degree $4$? It must involve a sum of fourth powers.
Thanks!