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I want to factorize $a^2(b − c)^3 + b^2(c − a)^3 + c^2(a − b)^3$ . By inspection , I can see that substituting $b$ for $a$ yields $0$ thus $(a-b)$ is a factor . Similarly $(c-a)$ and $(b-c)$ are factors . But I can't figure out other factors. Yes , I know that I can find it using long division , but there must be an easy way . Please help.

( Using wolfram alpha , I know that the factorization is $(a-b)(c-a)(b-c)(ab+ac+bc)$ , but I want to know how to figure that out . ( without long division ) )

Edit : see this page to understand what I mean - mathnerds - Is there any similar method for this question ?

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up vote 1 down vote accepted

E = The given expression, is homogeneous and cyclic in a, b and c, and is of degree 5.

p = (a - b)(b - c)(c - a) is cyclic of degree 3.

If E = p.q, then q should be homogeneous and cyclic in a, b and c, and is of degree 2.

The most general expression of q is then $m(a^2 + b^2 + c^2) + n(ab + bc + ca)$; for some m and n.

Thus, $E = (a - b)(b - c)(c - a)[m(a^2 + b^2 + c^2) + n(ab + bc + ca)]$

m and n can be obtained by comparing coefficients. [Subsequently, found the following is simpler. See also problem # 465127.]

m and n can also be obtained by substituting 2 different sets of values for a, b, and c.

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If for some reason you did not want to divide out the known terms, you could note that

  • The remaining factor has degree 2;
  • It is symmetric under permutations of the variables.

This tells you that the last factor is of the form $\alpha(a+b+c)^2 + \beta(a^2+b^2+c^2).$ Since variables do not appear to the fourth power in your original expression, $\alpha+\beta=0$ and then equating a single coefficient is enough to get you $\alpha=\frac{1}{2}, \beta=-\frac{1}{2}$.

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"Since variables do not appear to the fourth power in your original expression" - meaning no squares appear in the quotient. – anon Aug 11 '13 at 8:04

Taking into account what said before, perform the division and the end result is


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But isn't that division a big task ? ( without using wolfram alpha or similar resources ) Shouldn't there be an easier way ? – A Googler Aug 11 '13 at 7:50
@AGoogler That's already a very easy method... – Potato Aug 11 '13 at 7:53
Why have you factorized it like that if (a-b)(c-a)(b-c)(ab+ac+bc) looks better ? – A Googler Aug 11 '13 at 7:53
You have to expand out the first two terms as a cubic polynomial in $a$, and then perform polynomial division. Not something I could do in my head, but not really daunting, either. – user7530 Aug 11 '13 at 7:54

The remaining factor after dividing your polynomial $P$ by $Q = (a-b)(b-c)(c-a)$ is homogeneous of degree $2$ and is a symmetric polynomial in $a,b,c$ (because $P$ and $Q$ are antisymmetric), so it can be expressed in terms of the elementary symmetric polynomials: it must be $s (a + b + c)^2 + t (ab + bc + ac)$. Substitute $a=0,b=1,c=-1$ which makes $P = -2$, $Q = 2$, $a+b+c = 0$ and $ab + bc + ac= -1$, to get $t = 1$. Substitute $a=2,b=-3,c=-6$ with makes $P = 0$, $Q = -120$ and $ab + bc + ac = 0$ to get $s = 0$. So it must be $P/Q = ab + bc + ac$.

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But the factorization is (a−b)(c−a)(b−c)(ab+ac+bc) . Why did you got -(ab+ac+bc) [negative] instead of positive (ab+ac+bc) ? – A Googler Aug 11 '13 at 8:08
I think Substituting a=0,b=1,c=−1 makes Q=positive 2 not negative 2. – A Googler Aug 11 '13 at 8:12
Sorry, some wrong signs there. I'll edit. – Robert Israel Aug 11 '13 at 17:05

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