Question

Use euclid's division lemma to show that the cube of any positive integer is of the form 9m,9m +1,9m+8 brainly .in

Answer 1

Case 1
a= 3q
a³=(3q)³
=27q³
=9q(3q³)
=9m...........bcx (3q³) i s equal to m

case 2..
a=3q+1

a³=(3q+1)³
a³=27q³+27q²+9q+1
a³=9(3q³+3q²+1)+1
a³=9m+1
m =,(3q³+3q²+1)
.
case 3

a=3q+2
a³=(3q+2)³
a³=27q³+54q²+36q+2
a³=9(3q³+6q²+4q)+2
a³=9m +2

m=(3q³+6q²+4q)




HOPE ITS HELPFUL

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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