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.
$$a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)$$
$$=ab^2-ac^2+bc^2-a^2b+a^2c-cb^2$$
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 $1$.
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) 
=> 0+4-2=2m => 2m=2 =>m=1