Show that the difference of two consecutive cubes is never divisible by $3$. Here is my proof:
Let $n \in \Bbb Z$.  Then, $n$ is of the form $2k$(even) or $2k + 1$(odd), for some $k \in \Bbb Z$.
Without loss of generality (not sure if I can use this), let $n = 2k$.
Then, $n + 1 = 2k + 1$.
$$\begin{align}
(n + 1)^3 - n^3 & = (2k + 1)^3 - (2k)^3 \\
& = 8k^3 + 12k^2 + 6k + 1 - 8k^3 \\
& = 3(4k^2 + 2k) + 1
\end{align}$$
Let $m = (n + 1)^3 - n^3$.
Then, $m = 1 + 3(4k^2 + 2k) \Rightarrow m \equiv 1 \pmod 3$.
Therefore, $\forall m$, when divided by $3$, there remains a remainder $1$.  So $(n + 1)^3 - n^3$ is not divisible by $3$.
Do I also have to prove this for the other case where $n$ is odd, or is this proof sufficient?
 A: This can also be understood geometrically.
Arrange $n^3$ little cubes into a large cube of side length $n$, and then remove $(n-1)^3$ of them, leaving three flat sheets of cubes meeting at three edges.
Every cube left is either on the inside of one of the three sheets (and there are equally many of these on each sheet) or at one of the edges but not at the apex (again by symmetry each edge has equally many cubes in it), or it is the apex cube.
Therefore the total number of cubes left must be three times something, plus one cube at the apex.
A: $(n+1)^3 - n^3 = n^3 + 3n^2 + 3n + 1 - n^3  = 3(n^2 + n) + 1 \equiv 1$ (mod $3$)
A: Let, number 1 = x then 
     number 2 = x+1
(If Difference of cubes is divisible by 3 then if we mod that value by 3 it should give reminder as 0.)
Now, [Difference of cubes] mod 3 ,
= [(x+1)^3 - x^3] mod 3
= [x^3 + 3(x)(1)(x+1) + 1 - x^3] mod 3
= [3(x)(x+1) + 1] mod 3
= 1                         (because 3(x)(x+1) is always divisible by 3)

Thus, the reminder will be always 1.
A: The difference from one cube to the next has a pattern the cubes 1^3 thru 10^3 are  1;8;27;64;125;216;343;512;729;1000 the differences are 7;19;37;61;91;127;169;217;271 they are all a multiples of 3+1 because they increase by multiples of 6 which would be multiples of three as well  7+12=19+18=37+24=61+30=91+36=127+42=169+48=217+54=271.... So if you start with a number not divisible by 3 and you add a number that is divisible by 3 then you will get another number not divisible by 3 
