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I can do this problem by going over all the cases mod 7. Is there any other way to do it?. I think this way is too long , so I wonder if there is another way to do it or some way to simplify my solution.

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  • $\begingroup$ A simple modulo argument cannot be "way to long". $\endgroup$ – Dietrich Burde Nov 25 '17 at 9:45
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Modulo 9 the cubes are -1,0 or 1. The only combinations that work for your equation involve 0.

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According to Fermat's Last Theorem, your equation has no whole number solutions (besides $0$ of course).

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  • $\begingroup$ Certainly, but because of this, surely the OP was assuming that $x$, $y$ and $z$ are $3$-adic integers? $\endgroup$ – Angina Seng Nov 25 '17 at 8:21
  • $\begingroup$ @LordSharktheUnknown Possibly. But there would still be no rational solutions, right? $\endgroup$ – Badr B Nov 25 '17 at 20:24

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