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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 therefore 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 non zero . 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 how to find this coefficient.

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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.... – N. S. Feb 25 '13 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 $c$. – hardmath Feb 25 '13 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 $m = 1$ ;-) – vonbrand Feb 25 '13 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). – Calvin Lin Feb 25 '13 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 $1$.

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Nice Aditya. Also like your subsequent question. +1 for both – Shailesh Oct 14 '15 at 13:45
Thank you!.. :D – Aditya Agarwal Oct 14 '15 at 13:47

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