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Factor $ (a - b)^3 + (b - c)^3 + (c-a)^3$ by SYMMETRY.

Okay, this is the problem. Let $f(a) = (a - b)^3 + (b-c)^3 + (c-a)^3$ obviously, if you let $a = b$ then, $f(b) = 0$, thus $(a - b)$ is a factor of $f(a)$. Then someone said :

If $(a - b)$ is a factor then $(b - c)$ and $(a-c)$ must be factors as well by symmetry.

But $f(a, b, c)$ is not symmetric, actually, $f(b, a, c) = -f(a, b, c)$ it is an alternating polynomial, so what is up with the solution?

Amad27
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  • I see that in the quoted text, the words "by symmetry" occur twice. I do not see anyone claiming anywhere that $f(a,b,c)$ is a symmetric polynomial. – David K Oct 16 '15 at 21:40

2 Answers2

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The symmetry to use is not to change $f(a,b,c)$ to $f(b,a,c)$. Instead, change $f(a,b,c)$ to $f(b,c,a)$ or $f(c,a,b)$.

This is the same as replacing $(a - b)^3 + (b - c)^3 + (c - a)^3$ by either $(b - c)^3 + (c - a)^3 + (a - b)^3$ or $(c - a)^3 + (a - b)^3 + (b - c)^3$, which clearly makes no change in the value. So the same reasoning that showed $(a-b)$ is a factor must work for $(b-c)$ and $(c-a)$ as well.

And if $(c-a)$ is a factor, $(a - c)$ is a factor.

David K
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The function's symmetry doesn't extend to a, so $a-c$ is factor but not $b-c$.

Jan Stout
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