Question

Prove that cube of any positive integer of the form 9m,9m+1 or 9m+8 for some integer m.

Answer 1

m3 = 9m
m = root3m
m/root3 = m
3m = root3m
m = 3m
So m = 0

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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