Question

Use Euclid division lemma to show that the cube of any positive integers is of the from 9m.9m+1or9m+8

Answer 1

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

Answer 2

Question:-

By using Euclids Division lemma to show that cube of any positive integer is of the form of 9m, 9m +1 , 9m +8

★ Answer:−

let, A be any positive integer .

Then, B = 3

And q and R be some positive integer.

$\therefore{By \:using \:Euclids \:Division \:Lemma, we \:get}$

a = 3q + r ..... ( where , 0 <r <3)

. ° . The possible Values of R are - 0, 1 ,2

when , r = 0

↦a = 3q + r

↦️a = 3q + 0

↦️A = 3q

Now , cubing on both sides

↪️ a³ = (3q )³

↪️a³ = 27q³

↪️a³ = 9 × 3q ³

☆ a³ = 9m. { where, 3q³ = m is some integer }

when , r = 1

↦a = 3q +1

Now, cubing on both sides -

↪️a³ = ( 3q + 1 ) ³

↪️a³ = ( 3q ) ³ + 3 ×( 3q)² × 1 + 3× 3q (1 )² + (1)³

↪️a ³ = 27 q³ + 27q² + 9q + 1

↪️a³ = 9( 3q³ + 3q² + q ) + 1

☆ a³ = 9m + 1

( where, 3q³ + 3q² +q = m is some integer )

when r = 2

↦️a = 3q +2

Now, cubing on both sides

↪️a ³ = ( 3q + 2) ³

↪️a³ = ( 3q)³ + 3× (3q ) ² × 2 + 3× 3q × (2)²+(2)³

↪️a ³ = 27q³ + 54q² + 36q + 8

↪️a³ = 9 ( 3q³ + 6q² + 4q) + 8

️☆ a³ = 9m + 8

( where, 3q³ +6q² +4q = m is some integer)

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