Question

using Euclid's division lemma to show that the cube of any positive integer is of the form 9m,9m+1 or 9m+2, where m is some integer

Answer 1

Let a be any integer in the form of 3q,3q+1,3q+2

when a=3q c.b.s

Answer 2

Let 'x' be any +ev integer

by Euclids Lemma

x=bq+r , 0 <= r <b
Take , b=3
x = 3q,3q+1,3q+2.

x=3q
x^3=(3q)^3q
= 27q
= 9×(3q^3)
= 9m

x=3q+1
x^3=(3q+1)^3q
=(3q)^3+3×(3q)^2(1)+3×(3q)(1)^2+1^3
=27q^3+27q^2+9q+1
=9 (3q^3+3q^2+q)+1
=9m+1

x=3q+2
x^3=(3q+2)^3
=(3q)^3+3 (3q)^2 (2)+3 (3q)(2)^2+2^3
=27q^3+54q^2+36q+8
=9 (3q^3+6q^2+4q)+8
=9m+8

HENCE PROVED


I hope you understood.
mark it as Brainly