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I noticed the following amusing pattern when playing around in Wolfram Alpha:

$a(b-c) + b(c-a) + c(a-b) = 0$

$a^2(b-c) + b^2(c-a) + c^2(a-b) = -(a-b)(b-c)(c-a)$

$a^3(b-c) + b^3(c-a) + c^3(a-b) = -(a-b)(b-c)(c-a)(a+b+c)$

$a^4(b-c) + b^4(c-a) + c^4(a-b) = -(a-b)(b-c)(c-a)(a^2+b^2+c^2 + ab+ bc +ca)$

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$a^n(b-c) + b^n(c-a) + c^n(a-b) = -(a-b)(b-c)(c-a)\left(\displaystyle \sum_{i+j+k=n-2} a^ic^jb^k\right)$

I can prove that $(a-b)(b-c)(c-a)$ is a factor. How do we get the last factor? Is it always irreducible?

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    i would try induction if $n$ is a natural number – 2017-01-19
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    There should be negative signs before $(a-b)(b-c)(c-a) \ldots$ on the RHS. – 2017-01-19
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    @NgChungTak: You are right, thanks. – 2017-01-20

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Image of Solution

For irreducibility plz see https://mathoverflow.net/questions/98043/is-complete-homogeneous-symmetric-polynomials-an-irreducibile-element-in-polyno

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    Nice solution! Thanks! I did not think of using the determinant. – 2017-01-28