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I was solving the question that wanted to factor $a^4(b-c)+b^4(c-a)+c^4(a-b)$. My idea was to factor a $(c-a)$ in first step.So $$b(a^4-c^4)+ac(c^3-a^3)+b^4(c-a)=a^4(b-c)+b^4(c-a)+c^4(a-b)$$

$$(c-a)(-ba^3-bc^3-a^2bc-abc^2+ac^3+c^3a+a^2c^2+b^4)$$

And i was not able to continue this. I put the polynomial in Wolfram Alpha and the result was $$(a-b)(a-c)(b-c)(a^2+a b+a c+b^2+b c+c^2)$$

but it did not provided any steps.I want to know is there any better way to factor this polynomial than my first idea or a general method?Does this Factorization can be used in here? $$a^2(c-b)+b^2(a-c)+c^2(b-a)=(a-b)(b-c)(c-a)$$

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Well, here is one way. First note that setting $a=b$ makes the polynomial zero, so $(a-b)$ must be a factor. By symmetry, $(b-c),(c-a)$ are also factors.

Then by degree considerations you must have a quadratic left, which can be found by division or comparing coefficients.

Then attempt to factor the quadratic, (which in this case is futile) and conclude.

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  • $\begingroup$ thanks.is there a name for this method?i want to study a little bit about it. $\endgroup$ Sep 4, 2014 at 5:30
  • $\begingroup$ The closest is Factor Theorem (en.wikipedia.org/wiki/Factor_theorem). Useful for factoring polynomials/ determinants etc. $\endgroup$
    – Macavity
    Sep 4, 2014 at 5:36

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