Question

Using euclids division lemma show that the cube of any positive integer is either in the form of 9m or 9m 1 or 9m 8

Answer 1

Euclid's division lemma :

$\boxed{\boxed{a \:=\: bq \:+ \:r}}$

Here, a = any positive integer; which are in the form 3q, 3q + 1 and 3q + 2.

$\underline{Case \:1.}$

a = 3q

cube on both sides

(a)³ = (3q)³

a³ = 27q³

a³ = 9(3q³)

$\boxed{{a}^{3}\: = \:9m}$

[ As m = 3q³ ]

$\underline{Case\: 2.}$

a = 3q + 1

cube on both sides

(a)³ = (3q + 1)³

a³ = 27q³ + 27q² + 9q + 1

a³ = 9(3q³ + 3q² + q) + 1

$\boxed{{a}^{3} \:=\: 9m \:+ \:1}$

[ As m = 3q³ + 3q² + q ]

$\underline{Case \:3.}$

a = 3q + 2

cube on both sides

(a)³ = (3q + 2)³

a³ = 27q³ + 54q² + 36q + 8

a³ = 9(3q³ + 6q² + 4q) + 8

$\boxed{{a}^{3}\: =\: 9m\: + \:8}$

[ As m = 3q³ + 6q² + 4q ]

Hence;

Cube of any positive integer can be expressed in the form 9m, 9m + 1 or 9m + 8.

Answer 2

let us start by taking 'a' any positive integer and b = 3 (since any number which is divisible by 9, is also divisible by 3)

by Euclid's division algorithm,

a = bq + r where 0 ≤ r < b

therefore r = 0, 1 or 2

when r = 0, a = 3q + 0 ---------(i)

when r = 1, a = 3q + 1 ---------(ii)

when r = 2, a = 3q + 2 ---------(iii)

cubing both sides of (i), (ii), (iii)

➡ a³ = 3q³

= (9q)³

= 27q³

= 9(3q³)

= 9m

➡ a³ = (3q + 1)³

using identity, (a + b)³ = a³ + b³ + 3a²b + 3ab²

= (3q)³ + (1)³ + 3(3q)²(1) + 3(3q)(1)²

= 27q³ + 1 + 27q² + 9q

= (27q³ + 27q² + 9q) + 1

= 9(3q³ + 3q² + q) + 1

= 9m + 1

➡ a³ = (3q + 2)³

using identity, (a + b)³ = a³ + b³ + 3a²b + 3ab²

= (3q)³ + (2)³ + 3(3q)²(2) + 3(3q)(2)²

= 27q³ + 8 + 54q² + 36q

= (27q³ + 54q² + 36q) + 8

= 9(3q³ + 6q² + 4q) + 8

= 9m + 8

hence, it's proved that the cube of any positive integer is either in the form of 9m or 9m 1 or 9m 8.