Math, asked by 916006556193, 2 months ago

each of these and show that they
5. Use Euclid's division lemma to show that the cube of any positive integer is of the form
9m, 9m + 1 or 9m + 8.​

Answers

Answered by chaudharyashutosh39
1

Step-by-step explanation:

Let a and b be two positive integers, and a>b

a=(b×q)+r where q and r are positive integers and

0≤r<b

Let b=3 (If 9 is multiplied by 3 a perfect cube number is obtained)

a=3q+r where 0≤r<3

(i) if r=0,a=3q (ii) if r=1,a=3q+1 (iii) if r=2,a=3q+2

Consider, cubes of these

Case (i) a=3q

a

3

=(3q)

3

=27q

3

=9(3q

3

)=9m where m=3q

3

and 'm' is an integer.

Case (ii) a=3q+1

a

3

=(3q+1)

3

[(a+b)

3

=a

3

+b

3

+3a

2

b+3ab

2

]

=27q

3

+1+27q

2

+9q=27q

3

+27q

2

+9q+1

=9(3q

3

+3q

2

+q)+1=9m+1

where m=3q

3

+3q

2

+q and 'm' is an integer.

Case (iii) a=3q+2

a

3

=(3q+2)

3

=27q

3

+8+54q

2

+36q

=27q

3

−54q

2

+36q+8=9(3q

3

+6q

2

+4q)+8

9m+8, where m=3q

3

+6q

2

+4q and m is an integer.

∴ cube of any positive integer is either of the form 9m,9m+1 or 9m+8 for some integer m

Hence Proved.

I hope it will help you.

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