Question 462731
Every integer falls into one of the following modulo classes
7n
7n + 1
7n + 2
7n + 3
7n + 4
7n + 5
7n + 6

For the first class, {{{(7n)^3 = 0^3 = 0(mod7)}}}, which simply says that the cube of any multiple of 7 is still divisible by 7.

For the 2nd class 7n + 1,
{{{7n + 1 = -6 (mod7)}}} ==> {{{(7n+1)^3 = (-6)^3 = -216 = 1 (mod7)}}}.
For the 3rd class 7n + 2,
{{{7n + 2 = -5 (mod7)}}} ==> {{{(7n+2)^3 = (-5)^3 = -125 = 1 (mod7)}}}.
For the 4nd class 7n + 3,
{{{7n + 3 = -4 (mod7)}}} ==> {{{(7n+3)^3 = (-4)^3 = -64 = 6 (mod7)}}}.
For the 5th class 7n + 4,
{{{7n + 4 = -3 (mod7)}}} ==> {{{(7n+4)^3 = (-3)^3 = -27 = 1 (mod7)}}}.
For the 6th class 7n + 5,
{{{7n + 5 = -2 (mod7)}}} ==> {{{(7n+5)^3 = (-2)^3 = 8 = 1 (mod7)}}}.
For the 7th class 7n + 6,
{{{7n + 6 = -1 (mod7)}}} ==> {{{(7n+6)^3 = (-1)^3 = -1 = 6 (mod7)}}}.

In other words, the cube of any integer is either of the form 7k, 7k +1, or 7k + 6.  But numbers of the form 7k + 6 are the same numbers of the form 7k - 1.  The proof is complete.