Factoring $ab^3 - a^3 b + bc^3 - b^3 c + ca^3 - c^3 a$ 
Factor $$ab^3 - a^3 b + bc^3 - b^3 c + ca^3 - c^3 a$$

I used the factor theorem to get factors
$$f(a, b, c)=(a-b)(b-c)(a-c)\;g (a, b, c)$$
for some polynomial $g (a, b, c)$.
How can I continue using this method?
(sorry for the previously messed up question I'm new to this website and didn't fully understand the guidelines).
 A: From the degree of the original polynomial (4) and the degree of the factor that you found (3), the remaining factor has to be of degree 1. Since that factor has cyclic symmetry between $a$, $b$, and $c$, it must be a scalar multiple of  $a+b+c$. A trial valuation with, say, $a=0$, $b=1$, and $c=2$ will then give you the scaling number.
A: $$ab^3-a^3b+bc^3-b^3c+ac^3-a^3c$$ $$=ab(a+b)(b-a)+bc (b+c)(c-b)+ac (a+c)(a-c)$$ $$=ab( (a+b+c)-c)(b-a)+bc ((a+b+c)-a)(c-b)+ac  ((a+b+c)-b)(a-c)$$ $$=(a+b+c) [ab (b-a)+bc (c-b)+ac (a-c)]-abc (b-a+c-b+a-c)$$ $$=(a+b+c) [ab (b-a)+bc (c-b)+ac (a-c)]+0$$$$=(a+b+c)[(a-b)(b-c)(c-a)]$$
A: @JohnBentin's method is the slickest, but if you don't want to check one example to get the scaling factor, let $x:=b^2-a^2,\,y:=c^2-b^2$ so your sum is$$\begin{align}abx+bcy-ca(x+y)&=a(b-c)x+c(b-a)y\\&=a(b-c)(b-a)(b+a)+c(b-a)(c-b)(c+b)\\&=(b-a)(c-b)(c(c+b)-a(b+a))\\&=(b-a)(c-b)(c^2+bc-ab-a^2)\\&=(b-a)(c-b)(c-a)(a+b+c).\end{align}$$In fact, his method helps you guess some components of this calculation as you go, so it's not as taxing as it looks.
