The cube of any integer has one of the forms, 9k, 9k + 1, or 9k + 8.
Let n be an integer. By the Division Algorithm, either
n = 3m
n = 3m + 1
n = 3m + 2
If n = 3m, then n^3 = (3m)
3 = 27m^3 = 9 (3m^3)=9k, for k = 3m^3
If n = 3m + 1, then n3 = (3m + 1)^3 = 27m^3 + 27m^2 + 9m +1=
9 (3m^3 + 3m^2 + m)+1=9k + 1, for k = 3m^3 + 3m^2 + m
If n = 3m + 2, then n^3 = (3m + 2)^3 = 27m^3 + 54m^2 + 36m +8=
9 (2m^3 + 2m^2 + 4m)+8=9k + 8, for k = 2m^3 + 2m^2 + 4m
Hence, for any integer n, n3 has one of the forms, 9k, 9k + 1, or 9k + 8.
