Question

use euclid division lema to show that cube of any positive integeris either of the form 9m,9m+1 or 9m+8

Answer 1

take the numbers as 3q, 3q+1 and 3q+2 and cube them such that

x=bq+r where q>0 and 0<r<b

Answer 2

Hey mate your answer

Let a be any positive integer and b = 3

a = 3q + r, where q ≥ 0 and 0 ≤ r < 3

Therefore, every number can be represented as these three forms. There are three cases.

Case 1: When a = 3q,

Where m is an integer such that m =

Case 2: 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 3: 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)

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