Question

use Euclid division lemma to show that the cube of any positive integers is of the from 9m9m+8​

Answer 1

Answer:

let b=3

a=3q+r. (r=0,1,2)

a=3q+0

a=3q

a^3=(3q) ^3

a^3=27q^3

now take 9 common from RHS

a^3=9(3q^3)

{3q ^3} =m

a^3= 9m

〰️〰️〰️〰️

a=3q+1

a^3=(3q+1)^3

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

a^3=(27q^3+3*9q+9q*2)+1

take 9 common

a^3=9(3q^3+3q+2q)+1

a^3=9m+1

similarly take a=3m+2

take 2^3 in side

Answer 2

$\huge\underline\bold{Answer:-}$

let, A be any positive integer .

Then, B = 3

And q and R be some positive integer.

. ° . By using Euclids Division Lemma,we get

a = 3q + r ( where r = 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, we get

⇒️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