Express $-a^3 b + a^3 c+ab^3-a c^3-b^3c+bc^3$ as a product of linear factors. 
Express
  $$-a^3 b + a^3 c+ab^3-a c^3-b^3c+bc^3$$
  as a product of linear factors.

I have tried rewriting the expression as:
$$ab^3-a^3b + a^3c-ac^3 +bc^3-b^3c$$
$$= ab(b^2-a^2)+ac(a^2-c^2)+bc(c^2-b^2)$$
$$= ab(b-a)(b+a) + ac(a-c)(a+c)+bc(c-b)(c+b)$$
Which is at least a product of linear factors. 
However I am now stuck as to how to proceed.
Also I would prefer the $(b^2-a^2)$ and the $(c^2-b^2)$ factors of my second line of working to be in the form $(a^2-b^2)$ and $(b^2-c^2)$, for 'neatness' , if possible
 A: This is maybe not the most elegant answer, but I think it's relatively efficient. By trial and error we find that $a=b$, $a=c$ and $b=c$ make the expression vanish. Hence we write it as
$$\begin{align}
(a-b)(a-c)(b-c)f&=(a^2b-a^2c-ab^2+ac^2+b^2c-bc^2)f\\
&=-a^3 b + a^3 c+ab^3-a c^3-b^3c+bc^3
\end{align}$$
Now you could either divide one side by the other, or stare at it for a bit and note that $f=(a+b+c)$ makes it work. 
This tactic only works when the things you're doing is a homework exercise, since you can be pretty sure that the resulting factorisation will be nice and easy, so trying all 'easy relations' will get you quite far generally.
Note that since all coefficients are $1$, and you expect the factorisation to be easy, it wouldn't even be that far fetched to also just try $a=-b-c$ and see what comes out. This would have given you the solution without any calculations
A: The key is to look for symmetry, so write it as symmetrically as possible:$$a^3(c-b)+b^3(a-c) +c^3(b-a).$$This has cyclic symmetry; so, if it has linear factors, then they must be invariant under cyclic permutation of $a$, $b$, and $c$. The factors have to include negative signs, so try the simplest first: say $b-c$. Putting $b=c$ makes the first term zero, and it's easy to see that the remaining terms cancel too under this condition. So $b-c$ is a factor, and hence $c-a$ and $a-b$ must be too. The remaining factor has to be linear to make up the quartic; so it must be $\pm(a+b+c)$, which is the only cyclically symmetric linear factor possible (barring numerical multiples, because we don't have numbers other than $\pm1$). It's now easy to check whether the plus or minus sign should be taken.
A: Seeing it as a polynomial of one variable might help.
$$\begin{align}&-a^3b+a^3c+ab^3-ac^3-b^3c+bc^3\\&=(c-b)a^3+(b^3-c^3)a+bc^3-b^3c\\&=(c-b)a^3+(b-c)(b^2+bc+c^2)a+bc(c-b)(c+b)\\&=(c-b)(a^3-(b^2+bc+c^2)a+bc(c+b))\\&=(c-b)((-a+c)b^2+(-ac+c^2)b+a^3-ac^2)\\&=(c-b)((c-a)b^2+c(c-a)b+a(a-c)(a+c))\\&=(c-b)(c-a)(b^2+bc-a(a+c))\\&=(c-b)(c-a)((b-a)(b+a)+c(b-a))\\&=(c-b)(c-a)(b-a)(a+b+c)\\&=(a-b)(b-c)(c-a)(a+b+c)\end{align}$$
A: the resultat should be $$- \left( b-c \right)  \left( a-c \right)  \left( a-b \right)  \left( a
+c+b \right) 
$$
