# prove $ab\in A$, if $A=\{x^3+y^3+z^3-3xyz\mid x,y,z\in \mathbb Z\}$, $a,b\in A$

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

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 $\det (\mathcal{D}_{a,b,c})=a^3+b^3+c^3-3abc$
• 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? Jul 6, 2014 at 15:06
$$A=ax+by+cz$$ $$B=cx+ay+bz$$ $$C=bx+cy+az$$ and its easy to see this satisfies the equation given by OP.