How could I show that if $a$ is an integer, then $a^3 \equiv 1, 0, 6 \mod 7$? I've literally tried everything including proofs by cases. If I were to try a an odd integer, then I would get $8k^3 + 4k^2 + 8k^2 + 4k + 1$, which obviously couldn't convert to a $\mod 7$, and I'm honestly not sure what I can do! Does anyone here have any tips?
Note, this is a soft question.
 A: Any integer $a$ can be written as one of the following : $a= 7m, 7m+1,7m+2,7m+3,7m+4,7m+5,7m+6$, for some integer $m$. 
The cubic residues of each (and therefore all possible cubic residues) modulo $7$ is then  $0,1^3,2^3,3^3,4^3,5^3,6^3$.
For easier computation, you can rewrite the$7m+4,7m+5,7m+6$ as $7m-3,7m-2,7m-1$ since they are equivalent modulo 7.
A: There are various ways to do this: I'll present the bashy casework method, mostly because it's often the first one that comes to mind.  I know there are better ways to do this, but this is what you probably should immediately think to try.   
If $a \equiv 0 \bmod 7$, $a^3 \equiv 0 \bmod 7$.  
If $a \equiv 1 \bmod 7$, $a^3 \equiv 1 \bmod 7$.  
If $a \equiv 2 \bmod 7$, $a^3 \equiv 8 \equiv 1 \bmod 7$.  
If $a \equiv 3 \bmod 7$, $a^3 \equiv 27 \equiv 6 \bmod 7$.  
If $a \equiv 4 \equiv -3 \bmod 7$, $a^3 \equiv -6 \equiv 1 \bmod 7$.  
If $a \equiv 5 \equiv -2 \bmod 7$, $a^3 \equiv -1 \equiv 6 \bmod 7$.  
If $a \equiv 6 \equiv -1 \bmod 7$, $a^3 \equiv -1 \equiv 6 \bmod 7$.  
Note that this covers all the bases since an integer $a$ can only take on these $7$ values $\bmod \ 7$.  Thus we're done.  
