The document NCERT Solutions(Part- 1)- Cubes and Cube Roots Class 8 Notes | EduRev is a part of the Class 8 Course Mathematics (Maths) Class 8.

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**Question:** **Find the one’s digit of the cube of each of the following numbers.**

**(i) 3331 (ii) 8888 (iii) 149 (iv) 1005(v) 1024 (vi) 77 (vii) 5022 (viii) 53**

**Solution:**

Number | Number ending in | Unit’s digit in the cube | |

(i) (ii) (iii) (iv) (v) (vi) (vii) (viii) | 3331 8888 149 1005 1024 77 5022 53 | 1 8 9 5 4 7 2 3 | 1 2 9 5 4 3 8 7 |

**Note: **

I. There are only 4 numbers, less than 100 which are perfect cubes. They are: 1, 8, 27 and 64.

II. There are only 5 numbers between 100 and 1000 which are perfect cubes. These are: 125, 216, 343, 512 and 729.

III. The least 4 digit perfect cube is 1000.

**Question 1. ****Express the following numbers as the sum of odd numbers using the above pattern?**

**(a) 6 ^{3} (b) 8^{3} (c) 7**

**Solution: **(a) n = 6 and (n – 1) = 5

∴ We start with (6 * 5) + 1 = 31

We have

6^{3} = 31 + 33 + 35 + 37 + 39 + 41 = 216

**(b) **n = 8 and (n – 1) = 7

∴ We start with (8 * 7) + 1 = 57

We have

8^{3} = 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71 = 512

**(c) **n = 7 and (n – 1) = 6

∴ We start with (7 * 6) + 1 = 43

we have

7^{3} = 43 + 45 + 47 + 49 + 51 + 53 + 55 = 343

**Question 2. ****Consider the following pattern:**

**2 ^{3} – 1^{3} = 1 + 2 * 1 * 3**

3^{3} – 2^{3} = 1 + 3 * 2 * 3

4^{3} – 3^{3} = 1 + 4 * 3 * 3

**Using the above pattern, find the value of the following:**

**(i) 7 ^{3} – 6^{3} (ii) 12^{3} – 11^{3} (iii) 20^{3} – 19^{3} (iv) 51^{3} – 50**

**Solution: **

**(i)** 7^{3} – 6^{3} = 1 + 7 * 6 * 3

= 1 + 126 = 127

**(ii) **12^{3} – 11^{3} = 1 + 12 * 11 * 3

= 1 + 396 = 397

**(iii)** 20^{3 }– 19^{3} = 1 + 20 * 11 * 3

= 1 + 1140 = 1141

**(iv)** 51^{3} – 50^{3} = 1 + 51 * 50 * 3

= 1 + 7650 = 7651

**Note: **In the prime factorisation of any number, if each factor appears three times, then the number is a perfect cube.

**Question:** **Which of the following are perfect cubes?**

**1. 400 2. 3375 3. 8000 4. 15625 5. 9000 6. 6859 7. 2025 8. 10648**

**Solution:**

**1. **We have 400 = 2 * 2 * 2 * 2 * 5 * 5

∵ 2 * 5 * 5 remain after grouping in triples.

∴ 400 is not a perfect cube

**2.** We have 3375 = 3 * 3 * 3 * 5 * 5 * 5

∵ The prime factors appear in triples.

∴ 3375 is a perfect cube.

**3.** We have 8000 = 2 * 2 * 2 * 2 * 2 * 2 * 5 * 5 * 5

∵ The prime factors of 8000 can be grouped into triples and no factor is left over.

∴ 8000 is a perfect cube.

**4. **We have 15625 = 5 * 5 * 5 * 5 * 5 * 5

∵ The prime factors of 15625 can be grouped into triples and no factor is left over.

∴ 15625 is a perfect cube.

**5.** We have 9000 = 2 * 2 * 2 * 3 * 3 * 5 * 5 * 5

∵ The prime factors of 9000 cannot be grouped into triples (because 3 * 3 are left over).

∴ 9000 is not a perfect cube.

**6.** We have 6859 = 19 * 19 * 19

∵ The prime factors of 6859 can be grouped into triples and no factor is left over.

∴ 6859 is a perfect cube.

**7.** We have 2025 = 3 * 3 * 3 * 3 * 5 * 5

∵ We do not get triples of prime factors of 2025 and 3 * 5 * 5 are left over.

∴ 2025 is not a perfect cube.

**8. **We have 10648 = 2 * 2 * 2 * 11 * 11 * 11

∵ The prime factors of 10648 can be grouped into triples and no factor is left over.

∴ 10648 is a perfect cube.

**Question:** **Check which of the following are perfect cubes.**

**(i) 2700 (ii) 16000 (iii) 64000 (iv) 900 (v) 125000(vi) 36000 (vii) 21600 (viii) 10000 (ix) 27000000 (x) 1000**

**What pattern do you observe in these perfect cubes?**

**Solution:**

**(i)** We have 2700 = 2 * 2 * 3 * 3 * 3 * 5 * 5

We do not get complete triples of prime factors, i.e. 2 * 2 and 5 * 5 are left over.

∴ 2700 is not a perfect cube.

**(ii)** We have 1600 = 2 * 2 * 2 * 2 * 2 * 2 * 5 * 5

After grouping is 3’s we get 5 * 5 which is ungrouped in triples.

∴ 1600 is not a perfect cube.

**(iii)** We have 64000 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 5 * 5 * 5

Since, we get groups of triples.

∴ 64000 is a perfect cube.

**(iv)** We have 900 = 2 * 2 * 3 * 3 * 5 * 5

which are ungrouped in triples.

∴ 900 is not a perfect cube.

**(v) **We have 125000 = 2 * 2 * 2 * 5 * 5 * 5 * 5 * 5 * 5

As we get all the prime factors in the group of triples.

∴ 125000 is a perfect cube.

**(vi)** We have 36000 = 2 * 2 * 2 * 2 * 2 * 3 * 3 * 5 * 5 * 5

While grouping the prime factors of 36000 in triples, we are left over with 2 * 2 and 3 * 3.

∴ 36000 is not a perfect cube.

**(vii)** We have 21600 = 2 * 2 * 2 * 2 * 2 * 3 * 3 * 3 * 5 * 5

While grouping the prime factors of 21600 in triples, we are left with 2 * 2 and 5 * 5.

∴ 21600 is not a perfect cube.

**(viii) **We have 10000 = 2 * 2 * 2 * 2 * 5 * 5 * 5 * 5

While grouping the prime factors into triples, we are left over with 2 and 5.

∴ 10000 is not a perfect cube.

**(ix)** We have 27000000 = 2 * 2 * 2 * 2 * 2 * 2 * 3 * 3 * 3 * 5 * 5 * 5 * 5 * 5 * 5

Since, all the prime factors of 27000000 appear in groups of triples.

∴ 27000000 is a perfect cube.

**(x)** We have 1000 = 2 * 2 * 2 * 5 * 5 * 5

Since, all the prime factors of 1000 appear in groups of triples.

∴ 1000 is a perfect cube.

Now, in these perfect cubes, we can say that “the number of zeros at the end of a perfect cube must be 3 or a multiple of 3, failing which the number cannot be a perfect cube.”

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