Factorization of cyclic polynomial 
Factorize $$a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)$$ 

Since this is a cyclic polynomial, factors are also cyclic 
$$f(a) = a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)$$
$$f(b) = b(b^2-c^2)+b(c^2-b^2)+c(b^2-b^2) = 0 \Rightarrow a-b$$
is a factor of the given expression. Therefore, other factors are $(b-c)$ and $(c-a)$. The given expression may have a coefficient a constant factor which is nonzero. Let it be $m$.
$$\therefore a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2) = m(a-b)(b-c)(c-a)$$
Please guide further on how to find this coefficient.
 A: $$a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)$$
$$=ab^2-ac^2+bc^2-a^2b+a^2c-cb^2$$
Now the method to factor cyclic expressions is to arrange the expression with the highest powers of the first variable, i.e:
We take powers of $a$ $$a^2c-a^2b+ab^2-ac^2+bc^2-cb^2$$
$$=a^2(c-b)-a(c^2-b^2)+bc(c-b)$$
$$=(c-b)(a^2-ac-ab+bc)$$
Now we look for powers of $b$,: 
$$=(c-b)(bc-ab-ac+a^2)$$
$$=(c-b)(b(c-a)-a(c-a))$$
$$=(c-b)(c-a)(b-a)$$
$$=(a-b)(b-c)(c-a)$$
Thus, $m$ is $1$.
A: You can put arbitrary values in $f(a,b,c)$ and get the value of $m$
Let's put $a=1, b=2, c=0$ in  $$ a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2) = m(a-b)(b-c)(c-a)$$
we get, $m = 1$
A: \begin{align}
a (b^2-c^2)+b (c^2-a^2)+c (a^2-b^2)
&=ab^2-ac^2+bc^2-ba^2+ca^2-cb^2\\
&=a^2c-a^2b+ab^2-ac^2+bc^2-cb^2\\
&=a^2 (c-b)-a (c^2-b^2)+bc (c-b)\\
&=(c-b)(a^2-ac-ab+bc)\\
&=(c-b)(bc-ab-ac-a^2)\\
&=(c-b)(b (c-a)-a (c-a))\\
&=(c-b)(b-a)(c-a)\\
&=(a-b)(b-c)(c-a)
\end{align}
A: From your last line $$a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)=m(a-b)(b-c)(c-a);$$
putting $a=0$, $b=1$, $c=2$ we get
$$
0(1^2-2^2)+1(2^2-0^2)+2(0^2-1^2)=m(0-1)(1-2)(2-0)\\
\implies 0+4-2=2m \implies 2m=2 \implies m=1.
$$
