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By my previous questions here and here I have been inspired to ask about the factor theorem; the multivariable case of it.

So take $f(a, b, c) = (a-b)^3 + (b-c)^3 + (c-a)^3$

$f(a, a, c) = f(b, b, c) = f(a, c, c) = f(a, b, a) = f(a, b, b)$ these give factors: $(b - a), (a - b), (b - c), (c - a), (c - b)$, so which ones am I supposed to use?

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Amad27
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  • You mean $(c-a)$ instead of $(b-a)$. – Aravind Oct 16 '15 at 18:37
  • This type of theorem is possible if you can extract as many factors as the total degree. This often happens for determinants, eg: the determinant of the Vandermonde matrix (https://en.wikipedia.org/wiki/Vandermonde_matrix). – Aravind Oct 16 '15 at 18:39
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    Use all factors which are distinct. Note that factors which differ only by multiplication with a unit (in this case all real numbers are units) are not distinct. As it is a cubic, you shouldn't get more than three such factors. – Macavity Oct 17 '15 at 02:27
  • @Macavity, yes, but which distinct ones should I use? $(b-c)(b-a)(c-a)$ is distinct but it wont end up giving the right answer. – Amad27 Oct 17 '15 at 07:58
  • Of course it will give the right answer. All you need is a unit (in this case $-3$) in front. That scalar can always be found by checking any one coefficient, or testing for some non-zero value of $f$. – Macavity Oct 17 '15 at 08:03
  • @Macavity, huh. Very nice; is there any explanation for why we need distinct factors? (But overall, it works very good then!) – Amad27 Oct 17 '15 at 09:58

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