Prove that a cube of any integer is either of the form 9k or 9k+1

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.