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let $A=\{x^3+y^3+z^3-3xyz\mid x,y,z\in \mathbb {Z}\}$, prove that:

if $a,b\in A$, then $ab\in A$,

I think we must find $A,B,C$ such $$A^3+B^3+C^3-3ABC=(a^3+b^3+c^3-3abc)(x^3+y^3+z^3-3xyz)$$ where $A,B,C,a,b,c,x,y,z\in \mathbb Z$, but I can't find it.

I think this result is interesting, I hope someone can solve it.

Thank you

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Hint : Consider the matrix : $$\mathcal{D}_{a,b,c}=\begin{pmatrix} a&b&c\\ c&a&b\\ b&c&a\\ \end{pmatrix}$$ See, $$\mathcal{D}_{a,b,c}\times \mathcal{D}_{x,y,z}=\mathcal{D}_{p,q,r}$$ Now $\det (\mathcal{D}_{a,b,c})=a^3+b^3+c^3-3abc$

Looks like this hint solved the problem almost. :P

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  • $\begingroup$ How can you find this? can you explain? Thank you $\endgroup$ – math110 Jul 6 '14 at 15:02
  • $\begingroup$ The determinant is very well-known, I found this from memory. And multiplying these two matrices is not too hard, just tedious, which I give you as exercise. Now multiplying matrices of the type given yields another matrix of the same type. And each such matrix corresponds to an element of $A$ and taking determinant we get the $A,B,C$ you ask for. Is this explanation making it clear? $\endgroup$ – shadow10 Jul 6 '14 at 15:06
  • $\begingroup$ Sure, happy to help. A nice problem btw :) $\endgroup$ – shadow10 Jul 6 '14 at 15:09
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This answer is contained in the other, but by symmetry it is natural to write

$$A=ax+by+cz$$ $$B=cx+ay+bz$$ $$C=bx+cy+az$$ and its easy to see this satisfies the equation given by OP.

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