Is the Product \(abc(a^3-b^3)(b^3-c^3)(c^3-a^3)\) Divisible by 7?

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SUMMARY

The product \(abc(a^3-b^3)(b^3-c^3)(c^3-a^3)\) is proven to be divisible by 7 for any integers \(a\), \(b\), and \(c\). The discussion highlights the application of modular arithmetic, specifically focusing on the properties of cubic differences and the behavior of integers under modulo 7. Key arguments include the fact that at least one of the factors \(a\), \(b\), or \(c\) must be divisible by 7, ensuring the entire product is divisible by 7.

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Show that for integer a,b,c the product $abc(a^3-b^3)(b^3-c^3)(c^3-a^3)$ is divisible by 7
 
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kaliprasad said:
Show that for integer a,b,c the product $abc(a^3-b^3)(b^3-c^3)(c^3-a^3)$ is divisible by 7
If $a$, $b$, or $c$ is divisible by $7$, we are done.

Otherwise, each of $a^3$, $b^3$, $c^3$ is congruent to $\pm1\pmod{7}$. Therefore, at least two of these integers are congruent $\pmod7$, and the corresponding factor is divisible by $7$.
 

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