Question

use euclids division lemma to show that the cube of any positive integer is of the form of 9m,9m+1  or 9m+8

Answer 1

let x be any positive integer. then by euclid's division lemma 
x = 3q ,or 3q + 1, or 3q +2
so, x³ = (3q)³ 
          = 27q³
          = 9(3q³)
          =9m where m = 3q³
similarly, x³ = (3q +1)³
                   = 27q³ + 1³ + 3(3q²)(1) + 3(3q)(1)²
                   = 27q³ + 1 + 27q² + 9q
                   = 27q³ + 27q² + 9q + 1
                   = 9[q(3q² + 3q +1)] + 1
                   = 9m + 1 where m= q(3q² +3q + 1)
again, x³ = (3q +2)³
               = 27q³ + 2³ + 3(3q)²(2) + 3(3q)(2)²
               = 27q³ + 8 + 54q² + 36q
               = 27q³ + 54q² + 36q + 8
               = 9[q(3q² + 6q + 4)] + 8
               = 9m + 8 where m= q(3q² + 6q + 4)

Answer 2

Step-by-step explanation:

Let a be any positive integer and b = 3

a = 3q + r, where q ≥ 0 and 0 ≤ r < 3

∴ r = 0,1,2 .

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 = (3q +1) ³

a = 27q ³+ 27q ² + 9q + 1

a = 9(3q ³ + 3q ² + q) + 1

a = 9m + 1 [ Where m = 3q³ + 3q² + q ) .

Case 3: When a = 3q + 2,

a = (3q +2) ³

a = 27q³ + 54q² + 36q + 8

a = 9(3q³ + 6q² + 4q) + 8

a = 9m + 8

Where m is an integer such that m = (3q³ + 6q² + 4q)

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

Hence, it is proved .