Show that for any $n \in \mathbb{Z}, n^3$ is congruent to 0,1,-1 modulo 9. Having a little difficulties with this one. Tried thinking of going down the line of even/odd proofs, but couldn't get anywhere. 
 A: any integer has at least one of these representations: 
$$3k, 3k+1, 3k-1$$
in cubic power
$$27k^3, 27k^3 + 27k^2 + 9k + 1, 27k^3 - 27k^2 + 9k -1$$
all of them are either dividable by 9 or remaining = -1/+1.
A: Modulo 9 table:
$x\equiv 0 \bmod 9 \rightarrow x^2\equiv 0 \bmod 9 \rightarrow x^3\equiv 0 \bmod9$
$x\equiv 1 \bmod 9\rightarrow x^2\equiv 1 \bmod 9 \rightarrow x^3\equiv 1 \bmod9$
$x\equiv 2 \bmod 9\rightarrow x^2\equiv 4 \bmod 9 \rightarrow x^3\equiv 8 \bmod9$
$x\equiv 3 \bmod 9\rightarrow x^2\equiv 0 \bmod 9 \rightarrow x^3\equiv 0 \bmod9$
$x\equiv 4 \bmod 9\rightarrow x^2\equiv 7 \bmod 9 \rightarrow x^3\equiv 1 \bmod9$
$x\equiv 5 \bmod 9\rightarrow x^2\equiv 7 \bmod 9 \rightarrow x^3\equiv 8 \bmod9$
$x\equiv 6 \bmod 9\rightarrow x^2\equiv 0 \bmod 9 \rightarrow x^3\equiv 0 \bmod9$
$x\equiv 7 \bmod 9\rightarrow x^2\equiv 4 \bmod 9 \rightarrow x^3\equiv 1 \bmod9$
$x\equiv 8 \bmod 9\rightarrow x^2\equiv 1 \bmod 9 \rightarrow x^3\equiv 8 \bmod9$
alternatively:
$(3k)^3=27k^3$
$(3k+1)^3=27k^3+3*(9k^2)+3(3k)+1=9(3k^3+3k^2+k)+1,$
$(3k+2)^3=3k^3+3(2*9k^2)+3(4*3k)+8=9(3k^3+6k^2+4k)+8$
The modulo table allways works for proving these things, but the other one is nicer. since $8\equiv -1 \bmod9$ you do get what you want.
