Question

Question Use Euclid’s Division Lemma to show that the cube of any positive integer is either of the form 9m, 9m + 1 or 9m + 8.

Answer 1

Answer:

a=3q+r

let us consider b=3

r(0,1,2)

a=3q+r

put r=0

a=3q+0

acube =(3q) cube = acube= 27 q cube

acube = 9 (3q) cube

acube = 9 m ( m= 3qcube)

put r=1

a= 3q+1

acube = 3qcube +1

acube =( 3q+1) cube

27q cube + 27 q square + 9 q+ 1

9(3qcube + 3q square+ 9) +1

acube=9m+1

put r= 2

a= (3q+2) cube

acube = 27qcube +54q cube + 36q+8

9(3qcube + 6qcube+4q)+8

acube =9m+8(m+8=3qcube+6qcube+4q)

from given question in answer

hence we prove

Answer 2

Answer:

Step-by-step explanation:

Let a be a positive integer. Here , b= 3

By Euclid's division lemma

a = 3q + r ..,.(1). Where r is equal to or greater than 0 and less than 3

Therefore , r = 0, 1,2,

CASE 1 : When r = 0

a = 3q .....( From 1 )

a^3 = (3q)^3

a^3 = 27q^3 = 9 ( 3 q³ )

a³= 9 m where m= 3q³

CASE 2 : when r = 1

a = 3q +1 ( from 1 )

a³= ( 3q+1)³

a³= 9q³+1+27q²+9q

a³= 9 ( q³ + 3q²+q ) + 1

a³= 9m +1 where m = q³ + 3q²+q

CASE 3 : when r = 2

a= 3q+ 2 ( from 1 )

a³ = ( 3q +2)³

a³= 27q³+8+54q²+36q

a³=9 ( 3q³ + 6 q²+4q ) +8

a³= 9m+ 8 where m= 3q³ + 6 q²+4q

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