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Question

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


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Solution

Step 1: Prove that cube of any positive integer is of the form 9m

Euclid's division lemma states that if there are two positive integers a and b, then there exists unique integers q and r such that a=bq+r, 0r<b

Now suppose b=3, then 0r<3

So, possible values of r=0,1,2

For, r=0 the equation according to Euclid's division lemma is;

a=3q

On cubing both sides, we get;

a3=3q3

a3=27q3

a3=93q3

a3=9m, where m=3q3

Step 2: Prove that cube of any positive integer is of the form 9m+1

For, r=1 the equation according to Euclid's division lemma is;

a=3q+1

On cubing both sides, we get;

a3=3q+13

a3=27q3+1+27q2+9q [a+b3=a3+b3+3a2b+3ab2]

a3=27q3+27q2+9q+1

a3=93q3+3q2+q+1

a3=9m+1, where m=93q3+3q2+q

Step 3: Prove that cube of any positive integer is of the form 9m+8

For, r=2 the equation according to Euclid's division lemma is;

a=3q+2

On cubing both sides, we get;

a3=3q+23

a3=27q3+8+54q2+36q [a+b3=a3+b3+3a2b+3ab2]

a3=27q3+54q2+36q+8

a3=93q3+6q2+4q+8

a3=9m+8, where m=93q3+6q2+4q

Hence, it is proved that the cube of any positive integer is of the form 9m,9m+1 or 9m+8.


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