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

**Solution:**

Suppose that there is a positive integer ‘a’. By Euclid’s division lemma, we know that for positive integers a and b, there exist unique integers q and r, such that a = bq + r, 0 ≤ r < b

Let b = 3, then 0 ≤ r < 3, that is, r = 0 or 1 or 2 but it can’t be 3 because r is smaller than 3.

So, the possible values for a are 3q or 3q + 1 or 3q + 2.

Now, find the cube of all the possible values of a.

If q is any positive integer, then its cube (let’s call it “m”) will also be a positive integer.

Now, observe carefully that the cube of all the positive integers is either of form 9m or 9m + 1 or 9m + 1 for some integer m.

Let 'a' be any positive integer and q = 3. Then, a = 3q + r for some integer q ≥ 0 and 0 ≤ r < 3

Therefore, a = 3q or 3q + 1 or 3q + 2

Case I: When a = 3q

(a)³ = (3q)^{3} = 27q^{3}

= 9 (3q^{3})

= 9m, where m is an integer such that m = 3q^{3}

Case II: When a = 3q + 1

(a)^{3} = (3q + 1)^{3}

(a)^{3} = 27q^{3} + 27q^{2} + 9q + 1

(a)^{3} = 9(3q^{3} + 3q^{2} + q ) + 1

(a)^{3} = 9m + 1, where m is an integer such that m = 3q^{3} + 3q^{2} + q

Case III: When a = 3q + 2

(a)^{3} = (3q + 2)^{3}

(a)^{3} = 27q^{3} + 54q^{2} + 36q + 8

(a)^{3} = 9(3q^{3} + 6q^{2} + 4q) + 8

(a)^{3} = 9m + 8, where m is an integer such that m = 3q^{3 }+ 6q^{2} + 4q

Thus, we can see that the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 1

**Video Solution:**

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

NCERT Solutions Class 10 Maths - Chapter 1 Exercise 1.1 Question 5

**Summary:**

Using Euclid's division lemma, it can be proved that the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

**☛ Related Questions:**

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