Factor $ab^3-ac^3+bc^3-ba^3+ca^3-cb^3$ 
a) Use the remainder theorem to prove that $(a+b+c)$ is a factor of $(a^3+b^3+c^3-3abc)$ . Then find the other factor.
b) Hence factor $(ab^3-ac^3+bc^3-ba^3+ca^3-cb^3)$

So far I have managed to find the other remainder being $(a^2+b^2+c^2-ab-ac-bc)$ but I don't understand how to use this to hence factor the expression. 
Also I was reading a solution which factorised the expression without using part a) but how does the second last line turn into the last line?
\begin{align}&\color{white}=(bc^3−cb^3)−(ac^3−ca^3)+(ab^3−ba^3)\\
&=bc(c^2−b^2)−ac(c^2−a^2)+ab(b^2−a^2)\\
&=(c−b)(c+b)bc−ac(c−a)(c+a)+ab(b−a)(b+a)\\
&=−(a−c)(b−c)(a−b)(a+b+c)\end{align}
Thanks in advance

The polynomial
$$f(a,b,c)=ab^3-ac^3+bc^3-ba^3+ca^3-cb^3$$
is an alternating function of $a$, $b$, $c$. That is it changes sign
whenever you swap two variables:
$$f(a,b,c)=-f(b,a,c)=-f(c,b,a)=-f(a,c,b).$$
This means that $a-b$, $a-c$ and $b-c$ are all factors (if a polynomial
changes sign when you swap $a$ and $b$ then $a-b$ is a factor). Therefore
$$f(a,b,c)=(a-b)(a-c)(b-c)g(a,b,c)$$
for some polynomial $g(a,b,c)$. As $f$ is alternating, $g$ is symmetric:
$$g(a,b,c)=g(b,a,c)=g(c,b,a)=g(a,c,b).$$
It's clear that $g$ is homogeneous of $a$, $b$ and $c$ of degree one.
So $g(a,b,c)=k(a+b+c)$ for some constant $k$. One only needs to
compute one coefficient to see that $k=-1$. Therefore
$$f(a,b,c)=-(a-b)(a-c)(b-c)(a+b+c).$$
As a follow-up you might try to factorise
$$ab^4-ac^4+bc^4-ba^4+ca^4-cb^4$$
or
$$a^2b^3-a^2c^3+b^2c^3-b^2a^3+c^2a^3-c^2b^3.$$
