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.

We can solve these equations by using substitution method where you substitute one equation to the other and solve for the value of the other variable. We do as follows:

y = 2x^2
y = –3x −1

 2x^2 = –3x −1
 2x^2 +3x +1 = 0

x1= -0.5
x2 = -1
y1 = 0.5
y2 = 2