Question

Show that cube of any positive integer is of the form 9m 9m+1 9m+8 brainly

Answer 1

let be... any positive odd integer is "a"
a=bq+r where 0>r>b
now we will take b = 3 for short process if we take 9 there will be 9 cases and it would be so large my friend
now possible values for r = (+-)0,1,2
take b =3
a=3q+r
put 1st value of r that is 0
a=3q+0
a=3q
now we will do cubing on both sides
a^3 = (3q)^3
a^3 = 27q^3
a^3 = 9q(3q^2)
a=9m
where m= q(3q^2)

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 .

THANKS

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