Q. 2

# By Euclid Division Lemma show that the cube of any positive integral number is of the form 9q or 9q + 1 or 9q + 8, where q is an integral number.

Answer :

According to the Euclid’s Division Lemma,

Let a be any positive integer and b = 9

Then by Euclid’s Lemma we have,

a = 9q + r

Where r = 0, 1, 2, 3, 4, 5, 6, 7, 8 [since 0 ≤ r ≤ b] and here value of b is 9]

So all the possible forms of a are as follows:

9q

9q + 1

9q + 2

9q + 3

9q + 4

9q + 5

9q + 6

9q + 7

9q + 8

Now to find the cubes of these values we have the following expansion formula:

(a + b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3}

(9q + b)^{3} = 729q^{3} + 243q^{2}b + 27qb^{2} + b^{3}

Now when we divide the above equation by 9 we get the quotient as

81q^{3} + 27q^{2}b + 3qb^{2} and the remainder is b^{3}.

So we have to consider the value of b^{3}

So to put b = 0

We get 9m + 0 = 9m

So to put b = 1

We get 1^{3} = 1 so we get = 9m + 1

So to put b = 2

We get 2^{3} = 8 so we get = 9m + 8

So to put b = 3

We get 3^{3} = 27 and it is divisible by 9, so we get = 9m

So to put b = 4

We get 4^{3} = 64 and when divided by 9 we get 1 as remainder, so we get = 9m + 8

So to put b = 5

We get 5^{3} = 125 and when divided by 9 we get 8 as remainder, so we get = 9m + 8

So to put b = 6

We get 6^{3} = 216 and when divided by 9 we get 0 as remainder, so we get = 9m

So to put b = 7

We get 7^{3} = 343 and when divided by 9 we get 1 as remainder, so we get = 9m + 1

So to put b = 8

We get 8^{3} = 512 and when divided by 9 we get 8 as remainder, so we get = 9m + 8

So it is proved that all the values are in the form of 9m, 9m + 1 and 9m + 8.

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