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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.

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    $\begingroup$ 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.... $\endgroup$
    – N. S.
    Feb 25, 2013 at 17:48
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    $\begingroup$ 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 $c$. $\endgroup$
    – hardmath
    Feb 25, 2013 at 18:00
  • $\begingroup$ 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 $m = 1$ ;-) $\endgroup$
    – vonbrand
    Feb 25, 2013 at 19:07
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    $\begingroup$ 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). $\endgroup$
    – Calvin Lin
    Feb 25, 2013 at 23:35

4 Answers 4

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

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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, $m = 1$

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  • $\begingroup$ Why does putting 'a=1, b=2, c=0' work? Furthermore, can it be done by putting any three consecutive integers? Can it be done by putting any three integers? $\endgroup$
    – Kawrno
    Aug 12, 2019 at 18:32
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\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}

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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 $$ 0(1^2-2^2)+1(2^2-0^2)+2(0^2-1^2)=m(0-1)(1-2)(2-0)\\ \implies 0+4-2=2m \implies 2m=2 \implies m=1. $$

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