<s>[INST] 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
 [/INST] Hint (without using the first part): considering it as a $3^{rd}$ degree polynomial in $a\,$:
$$ab^3-ac^3+bc^3-ba^3+ca^3-cb^3 = (c-b)\,a^3+(b^3-c^3)\,a+bc\,(b^2-c^2)$$
It can be easily verified that it has $a=b$ and $a=c$ as roots. Since the term in $a^2$ is missing, the sum of all three roots is $0$ by Vieta's relations, so the third root is $a=-(b+c)\,$. Therefore the polynomial is divisible by $(a-b)(a-c)(a+b+c)\,$. Comparing the leading coefficients, it follows that the scalar multiplier is $c-b\,$, so in the end:
$$ab^3-ac^3+bc^3-ba^3+ca^3-cb^3 = (c-b)(a-b)(a-c)(a+b+c)$$
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