Question

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

Answer 1

Let a be any positive integer
And b=3
Using Euclid division lemma
a = 3q+r ; 0<=3<3
Possible values of r = 0,1,2
Case 1- r =0
a=3q+r
a=3q+0
a³ = (3q)³
=27q³
= 9(3q)³
=9m (m = 3q³)
Case 2- r=1
a=3q+1
(a)³=(3q+1) ³
= (3q)³+1³+3*(3q)²*1 +3*3q+1²
= 27q³+ 27q²+9q+1
= 9(3q³+3q²+3)+1
=9m+1 (m=3q³+3q²+3)

Case 3- r=2
a=3q+2
a³= (3q+2)³
a = (3q)³+2³+3*(3q)²*2+3*3q*2²
= 27q³+8+54q²+36q
= 27q³+54q²+36q+8
= 9(3q³+6q²+4q)+8
= 9m+8 (m= 3q³+6q²+4q)
The cube of any positive integer is of the former 9m, 9m+1, 9m+8.

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 .