Question

to show cube of any positive integer is of the form 9 m 9 m + 1 or 9m+ 8

Answer 1

Let x be any positive integer.
Then x can be 3q, 3q+1, 3q+2. ( For some integer q )

CASE 1 :-
x = 3q
=> x^3 = (3q)^3
=> x^3 = 27q^3
=> x^3 = 9(3q^3)
=> x^3 = 9m. ( For 3q^3 = m). _______________(1)


CASE 2 :-
x = 3q+1
=> x^3 = (3q+1)^3
=> x^3 = 27q^3 + 27q^2 + 9q +1
=> x^3 = 9 ( 3q^3 + 3q^2 + q ) + 1
=> x^3 = 9m + 1. ( For 3q^3 + 3q^2 + q = m) ______(2)

CASE 3 :-
x = 3q+2
=> x^3 = (3q+2)^3
=> x^3 = 27q^3 + 54q^2 + 36q + 8
=> x^3 = 9 ( 3q^3 + 6q^2 + 4q ) + 8
=> x^3 = 9m + 8. ( For 3q^3 + 6q^2 + 4q = m) _____(3)

From equation (1), (2) & (3) we get,
Cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

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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