Question

show that cube of any positive integer is of the form 9M, 9m + 1 or 9m + 8 for some integer m

Answer 1

let "a" be a real no. divided by 3.
by euclid's division lemma, we get a=3q+r, 0_<r<3 , where q is a positive integer.
if,r=0
a=3q
a^3=(3q)^3
      =9*(3q^3)
      =9m
where m = 3q^3

now if , r=1
a=3q+1
a^3=(3q+1)^3
      =(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
m= 3q^3+3q^2+q

now if , r=2
a=3q+2
a^3=(3q+2)^3
      =(3q)^3+3.(3q)^2.2+3.3q.(2)^3
      =27q^3+54q^2+36q+8
      =9(3q^3+6q^2+4q)+8
m=3q^3+6q^2+4q

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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