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

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.
