Question

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

Answer 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 n3 = (3m)3 = 27m3 = 9 (3m3 ) = 9k, for k = 3m3

If n = 3m + 1, then n^3 = (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, n^3 has one of the forms, 9k, 9k + 1, or9k + 8.

Answer 2

yup