Step-by-step explanation:

To prove that,

The cube of any positive integer is of the form 9m, 9m+1 or 9m+8.

First of all,

LET's ASSUME that a be any positive integer and b = 3.

Therefore, we can write,

=> a = 3q + r,

Where q ≥ 0 and 0 ≤ r < 3.

Therefore, we have possible values of r = 0,1,2

Now, we know that,

Every number can be represented as these three forms.

Therefore, we have 3 cases.

CASE I: When a = 3q, 

=> a^3 = (3q)^3

=> a^3 = 27q^3

=> a^3 = 9(3q^3)

=> a^3 = 9m

Where, m is an integer such that m = 3q^3.

Case II: When a = 3q + 1,

=> a^ 3 = (3q +1)^ 3 

=>a^3 = 27q^ 3 + 27q ^2 + 9q + 1 

=> a^ 3 = 9(3q^ 3 + 3q ^2 + q) + 1

=> a ^3 = 9m + 1 

Where, m is an integer such that m = (3q ^3 + 3q^ 2+ q). 

Case III: When a = 3q + 2,

=> a^3 = (3q +2)^ 3 

=> a^ 3 = 27q^ 3 + 54q ^2 + 36q + 8 

=> a^ 3 = 9(3q ^3 + 6q ^2 + 4q) + 8

=> a^ 3 = 9m + 8

Where, m is an integer such that m = (3q^ 3 + 6q^ 2+ 4q). 

Hence, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

Thus, Proved.