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.

• The coefficient of $a^2b$ on the LHS is $-1$, while on the RHS $a^2b$ appears as $m(a)(b)(-a)$, thus the coefficient is $-m$. hence $-m=-1$... Note that you should make it clear why $m$ must be a number, using degrees.... Feb 25, 2013 at 17:48
• It's easy enough to factor out $(a-b)$: $a(b^2-c^2) + b(c^2-a^2) + c(a^2-b^2) = (a-b)(-ab-c^2+c(a+b))$. Then factor the second factor as a quadratic in $c$. Feb 25, 2013 at 18:00
• Ask Maxima to solve $a (b^2 - c^2) + b (c^2 - a^2) + c (a^2 - b^2) = m (a - b) (b - c) (c - a)$ and you get $m = 1$ ;-) Feb 25, 2013 at 19:07
• Another way to calculate (m) is to substitute in 3 distinct values of (a, b, c). For example, we could use ( a = 0, b = 1, c = 2 ), which gives a linear equation in (m). Feb 25, 2013 at 23:35

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