# Cubes And Cube Roots : Exercise 7.1 - Class 8 Maths - NCERT Solution

**Chapter 07 CUBES AND CUBE ROOTS**

**Exercise 7.1**

**Q.1 Which of the following numbers are not perfect cubes? **

**(i) 216 (ii) 128 (iii) 1000 (iv) 100 **

**(v) 46656**

** Sol.** (i) 216

2 | 216 |

2 | 108 |

2 | 54 |

3 | 27 |

3 | 9 |

3 | 3 |

1 |

Prime factors of 216 = 2 x 2 x 2 x 3 x 3 x 3

Here, each prime factor is appearing in multiple of 3.

Hence, 216 is a perfect cube.

(ii) 128

2 | 128 |

2 | 64 |

2 | 32 |

2 | 16 |

2 | 8 |

2 | 4 |

2 | 2 |

1 |

Prime factors of 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2

Here, each prime factor is not appearing in multiple of 3.

Hence, 128 is not a perfect cube.

(iii) 1000

2 | 1000 |

2 | 500 |

2 | 250 |

5 | 125 |

5 | 25 |

5 | 5 |

1 |

Prime factors of 1000 = 2 x 2 x 2 x 5 x 5 x 5

Here, each prime factor is appearing in multiple of 3.

Hence, 1000 is a perfect cube.

(iv) 100

2 | 100 |

2 | 50 |

5 | 25 |

5 | 5 |

1 |

Prime factors of 1000 = 2 x 2 x 5 x 5

Here, each prime factor is not appearing in multiple of 3.

Hence, 100 is not a perfect cube.

(v) 46656

2 | 46656 |

2 | 23328 |

2 | 11664 |

2 | 5832 |

2 | 2916 |

2 | 1458 |

3 | 729 |

3 | 243 |

3 | 81 |

3 | 27 |

3 | 9 |

3 | 3 |

1 |

Prime factors of 128 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3

Here, each prime factor is appearing in multiple of 3.

Hence, 46656 is a perfect cube.

**Q.2 Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube. **

**(i) 243 (ii) 256 (iii) 72 (iv) 6750 **

**(v) 100**

** Sol.** (i) 243

3 | 243 |

3 | 81 |

3 | 27 |

3 | 9 |

3 | 3 |

1 |

Prime factors of 243 = 3 x 3 x 3 x 3 x 3

Here, prime factor â€˜3â€™ is not appearing in triplet. Hence, to make 243 a cube, one more â€˜3â€™ needs to be multiplied.

Therefore, 243 x 3 = 3 x 3 x 3 x 3 x 3 x 3 = 729 is a perfect cube.

Hence, the smallest number by which 243 must be multiplied to obtain a perfect cube is 3.

(ii) 256

2 | 256 |

2 | 128 |

2 | 64 |

2 | 32 |

2 | 16 |

2 | 8 |

2 | 4 |

2 | 2 |

1 |

Prime factors of 256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

Here, prime factor â€˜2â€™ is not appearing in triplet. Hence, to make 243 a cube, one more â€˜2â€™ needs to be multiplied.

Therefore, 256 x 2 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 512 is a perfect cube.

Hence, the smallest number by which 256 must be multiplied to obtain a perfect cube is 2.

(iii) 72

2 | 72 |

2 | 36 |

2 | 18 |

3 | 9 |

3 | 3 |

1 |

Prime factors of 72 = 2 x 2 x 2 x 3 x 3

Here, prime factor â€˜3â€™ is not appearing in triplet. Hence, to make 72 a cube, one more â€˜3â€™ needs to be multiplied.

Therefore, 72 x 3 = 2 x 2 x 2 x 3 x 3 x 3 = 216 is a perfect cube.

Hence, the smallest number by which 72 must be multiplied to obtain a perfect cube is 3.

(iv) 675

3 | 675 |

3 | 225 |

3 | 75 |

5 | 25 |

5 | 5 |

1 |

Prime factors of 675 = 3 x 3 x 3 x 5 x 5

Here, prime factor â€˜5â€™ is not appearing in triplet. Hence, to make 675 a cube, one more â€˜5â€™ needs to be multiplied.

Therefore, 675 x 5 = 3 x 3 x 3 x 5 x 5 x 5 = 3375 is a perfect cube.

Hence, the smallest number by which 675 must be multiplied to obtain a perfect cube is 5.

(v) 100

2 | 100 |

2 | 50 |

5 | 25 |

5 | 5 |

1 |

Prime factors of 100 = 2 x 2 x 5 x 5

Here, prime factor â€˜2â€™ and â€˜5â€™ are not appearing in triplet. Hence, to make 100 a cube, one more â€˜2â€™ and â€˜5â€™ needs to be multiplied.

Therefore, 100 x 2 x 5 = 2 x 2 x 2 x 5 x 5 x 5 = 1000 is a perfect cube.

Hence, the smallest number by which 100 must be multiplied to obtain a perfect cube is 2 x 5 = 10.

**Q.3 Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube. **

**(i) 81 (ii) 128 (iii) 135 (iv) 192 **

**(v) 704**

** Sol.** (i) 81

3 | 81 |

3 | 27 |

3 | 9 |

3 | 3 |

1 |

Prime factors of 81 = 3 x 3 x 3 x 3

Here, prime factor â€˜3â€™ is not appearing in triplet. Hence, to make 243 a cube, one â€˜3â€™ needs to be divided.

Therefore, 81 Ã· 3 = 3 x 3 x 3 = 27 is a perfect cube.

Hence, the smallest number by which 81 must be divided to obtain a perfect cube is 3.

(ii) 128

2 | 128 |

2 | 64 |

2 | 32 |

2 | 16 |

2 | 8 |

2 | 4 |

2 | 2 |

1 |

Prime factors of 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2

Here, prime factor â€˜2â€™ is not appearing in triplet. Hence, to make 128 a cube, one â€˜2â€™ needs to be divided.

Therefore, 128 Ã· 2 = 2 x 2 x 2 x 2 x 2 x 2 = 64 is a perfect cube.

Hence, the smallest number by which 128 must be divided to obtain a perfect cube is 2.

(iii) 135

3 | 135 |

3 | 45 |

3 | 15 |

5 | 5 |

1 |

Prime factors of 135 = 3 x 3 x 3 x 5

Here, prime factor â€˜5â€™ is not appearing in triplet. Hence, to make 135 a cube, one â€˜5â€™ needs to be divided.

Therefore, 135 Ã· 5 = 3 x 3 x 3 = 27 is a perfect cube.

Hence, the smallest number by which 135 must be divided to obtain a perfect cube is 5.

(iv) 192

2 | 192 |

2 | 96 |

2 | 48 |

2 | 24 |

2 | 12 |

2 | 6 |

3 | 3 |

1 |

Prime factors of 192 = 2 x 2 x 2 x 2 x 2 x 2 x 3

Here, prime factor â€˜3â€™ is not appearing in triplet. Hence, to make 192 a cube, one â€˜3â€™ needs to be divided.

Therefore, 192 Ã· 3 = 2 x 2 x 2 x 2 x 2 x 2 = 64 is a perfect cube.

Hence, the smallest number by which 192 must be divided to obtain a perfect cube is 3.

(v) 704

2 | 704 |

2 | 352 |

2 | 176 |

2 | 88 |

2 | 44 |

2 | 22 |

11 | 11 |

1 |

Prime factors of 704 = 2 x 2 x 2 x 2 x 2 x 2 x 11

Here, prime factor â€˜11â€™ is not appearing in triplet. Hence, to make 704 a cube, one â€˜11â€™ needs to be divided.

Therefore, 704 Ã· 11 = 2 x 2 x 2 x 2 x 2 x 2 = 64 is a perfect cube.

Hence, the smallest number by which 704 must be divided to obtain a perfect cube is 11.

**Q.4 Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?**

** Sol.** Given, sides of plasticine are 5 cm, 2 cm, 5 cm.

We know that, volume = 5 cm x 2 cm x 5 cm = (5 x 5 x 2) cm3.

Here, two 5s and one 2 are which are not in a triplet.

Hence, multiplying this by 2 x 2 x 5 = 20, then it will be a perfect cube.

Thus, (5 x 5 x 2 x 2 x 2 x 5) = (5 x 5 x 5 x 2 x 2 x 2) = 1000 is a perfect cube.

Hence, 20 cuboids will be required to form a cube.