Question

Use Euclid’s division lemma to show that cube of any positive integer is either of the form
9q, 9q+1 or 9q + 8 for some integer ‘q’

Answer 1

Let x = 3q be any positive integer. Compare it with , a= bq+r
Here, b=3
since 0<r<b
therefore 0<r<3
then possible remainders are 0,1 and 2
That is x can be 3q or 3q+1 or 3q+2
When x=3q
cubing both sides, we get
x2= 27 we=9(3q3)=9m

similarly by cubing the other two remainders , u will got ur answer.

Answer 2

Step-by-step explanation:

[ note :- I am taking 'm' as some integer ]

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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