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Can anyone help me explain why if $x$ is an integer, then $(x^3+1)\bmod 3 = (x+1)^3 \bmod 3$? I know there are 3 cases. $x=0\bmod3,\ x=1\bmod3,$ and $x=2\bmod3$ totally new to this form of mathematics, could anyone help me setup the proof?

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    $\begingroup$ Consider expanding $(x+1)^3$. $\endgroup$ – Andrew T. Mar 2 '15 at 6:31
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    $\begingroup$ You could use the cases approach mentioned in the post. Or else you could note that $(x+1)^3=x^3+3x^2+3x+1$, so $(x+1)^3$ has the same remainder on division by $3$ as $x^3+1$. $\endgroup$ – André Nicolas Mar 2 '15 at 6:33
  • $\begingroup$ @BRIC-Fan: I don't quite get your edit. Are you seeing the equations as congruences or binary mods (=remainder operations)? Right now it is a mixture. For example you use equality signs even though your TeX (using pmod) suggests that a congruence was intended. At least in the first equation where binary mod (\bmod) should be used. I do speak (loudly) against ever using binary mod myself, so this is kinda kettle calling pot, but..? Ok, the original post wasn't entirely clear either :-) $\endgroup$ – Jyrki Lahtonen Mar 2 '15 at 7:14
  • $\begingroup$ @JyrkiLahtonen: I see it as a congruence or mod and not really an equation. But it can be viewed as both. Please correct if I am wrong as I am quite a new-b with mathjax. $\endgroup$ – DeepSea Mar 2 '15 at 7:17
  • $\begingroup$ aeon1: Judging from the title of your question you may be using mod as a remainder operation. However, your syntax is not consistent thoughout the post, and at some points it looks like you might intend a congruence instead. Please study this Wikipedia page that explains the differences, and explain in your own words how you see it. The reason I'm asking is that your interpretation has an impact on how to the equations/congruences should best be typeset. It is also IMO beneficial to you to be aware of the differences! $\endgroup$ – Jyrki Lahtonen Mar 2 '15 at 8:15
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If you're allowed to use it, by Fermat's Little Theorem $n^3\equiv n\pmod3$. Using this gives

$$x^3+1\equiv(x+1)^3\equiv x+1\pmod 3$$

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hint: $3x^2 = 0\pmod 3, 3x = 0 \pmod 3$

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$(x+1)^3 -(x^3+1) = 3(x+x^2)$.

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