Question

use euclids division lemma to show that the cube of any positive integer is of the form of 9m,9m+1  or 9m+8

Answer 1

a=bq+r
a>b
0$\leq$r<b

required remainders are:
0,1,2
a=3q+0.
a=3q+1.
a=3q+2.

1 case:      a=3q
             cubing ;  both ;  sides

a³ = 27q³.
    =9(3q³)
    ==9m
where m = 3q³.

2 case:    a=3q+1
          cubing  ;  both  ;  sides
a³=(3q+1 )³
=( 3q)³ +(1)³+3*3q*1 ( 3q+1)
=27q³+27q²+9q+1
=9( 3q³+3q²+q)+1
==9m+1
where m=3q³+3q²+q

3 case:   a=3q+2
           cubing  ;  both  ;  sides
a³=(3q+2)³
=(3q)³ + (2)³ + 3*3q*2( 3q+2)
=27q³+54q²+36q+8
=9( 3q³+6q²+4q)+8
==9m+8
where m =3q³+6q²+4q

therefore, cubes of any positive integers is either in the form of  9m, 9m+1, 9m+8 for integer 'm'.

Answer 2

Step-by-step explanation:

Let a be any positive integer and b = 3

a = 3q + r, where q ≥ 0 and 0 ≤ r < 3

∴ r = 0,1,2 .  

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 = (3q +1) ³  

a = 27q ³+ 27q ² + 9q + 1  

a = 9(3q ³ + 3q ² + q) + 1

a = 9m + 1  [ Where m = 3q³ + 3q² + q ) .

Case 3: When a = 3q + 2,

a = (3q +2) ³  

a = 27q³ + 54q² + 36q + 8  

a = 9(3q³ + 6q² + 4q) + 8

a = 9m + 8

Where m is an integer such that m = (3q³ + 6q² + 4q)  

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

Hence, it is proved .

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