Question

5. Use Euclid,s division lemma to show that the cube of any positive integer is of the form 9m, 9m+1 or 9m+8​

Answer 1

Step-by-step explanation:

Let a be any positive integer and b = 3

=> a=3q+r, where

$q \geqslant 0 \: and \: 0 \leqslant r < 3$

Using r as 0,1,2, every no. can be represented as these three forms. Three cases arise:-

Case 1 :- a = 3q

a³ = (3q)³

a³= 27q³

a³= 9m ( where m = 3q)

Case 2:- a = 3q+1

a³= (3q+1)³

a³=27q³+27q²+9q+1

a³=9(3q³+3q²+3q)+1

a³= 9m+1 (where m= 3q³+3q²+3q)

Case 3:- a = 3q+2

a³= (3q+2)³

a³=27q³+54q²+36q+8

a³=9(3q³+6q²+4q)+8

a³=9m+8 (where m= 3q³+6q²+4q)

Therefore, a³ can be represented as 9m , 9m+1, 9m+8.

Hence Proved...