## Cube and Cube Roots

###
*(NCERT Ex 7.1)*

**Q1: Which of the following numbers are not perfect cubes?**

(i) 216

(ii) 128

(iii) 1000

(iv) 100

(v) 46656

(i) 216

(ii) 128

(iii) 1000

(iv) 100

(v) 46656

Answer:

(i) The prime factorisation of 216 is as follows:

2 | 216 |

2 | 108 |

2 | 54 |

3 | 27 |

3 | 9 |

3 | 3 |

1 |

∴ 216 =

__2 ☓ 2 ☓ 2__☓

__3 ☓ 3 ☓ 3__= 2

^{3}☓ 3

^{3}

Here, each prime factor appears as many times as a perfect multiple of 3, therefore, 216 is a

perfect cube.

(ii) The prime factorisation of 128 is as follows:

2 | 128 |

2 | 64 |

2 | 32 |

2 | 16 |

2 | 8 |

2 | 4 |

2 | 2 |

1 |

∴ 128 =

__2 ☓ 2 ☓ 2__☓

__2 ☓ 2 ☓ 2__☓ 2

Here, each prime factor does not appear as many times as a perfect multiple of 3. Therefore, 128 is not a perfect cube.

(iii)The prime factorisation of 1000 is as follows.

2 | 1000 |

2 | 500 |

2 | 250 |

5 | 125 |

5 | 25 |

5 | 5 |

1 |

∴ 1000 =

__2 ☓ 2 ☓ 2__☓

__5 ☓ 5 ☓ 5__= 2

^{3}☓ 5

^{3}

Here, as each prime factor appears as many times as a perfect multiple of 3, therefore, 1000 is a perfect cube.

(iv)The prime factorisation of 100 is as follows.

2 | 100 |

2 | 50 |

5 | 25 |

5 | 5 |

1 |

∴ 100 = 2 ☓ 2 ☓ 5 ☓ 5

Here, each prime factor does not form triplet group(s). Therefore, 100 is not a perfect cube.

(iv) The prime factorisation of 46656 is as follows.

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 |

46656 =

__2 ☓ 2 ☓ 2__☓

__2 ☓ 2 ☓ 2__☓

__3 ☓ 3 ☓ 3__☓

__3 ☓ 3 ☓ 3__

= 2

^{3}☓ 2

^{3 }☓ 3

^{3 }☓ 3

^{3}

Here, each prime factor appears as many times as a perfect multiple of 3,

∴ 46656 is a perfect cube.

**Q2: 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) 675

(v) 100

(ii) 256

(iii) 72

(iv) 675

(v) 100

Answer:

**(i)**243 =

__3 ☓ 3 ☓ 3__☓ 3 ☓ 3

Here, two 3s are left which require one more 3 to form a triplet.

i.e. 243 ☓ 3 =

__3 ☓ 3 ☓ 3__☓

__3 ☓ 3 ☓ 3__= 729 is a perfect cube.

∴ the smallest natural number by which 243 should be multiplied to make it a perfect cube is 3.

**(ii)**256 =

__2 ☓ 2 ☓ 2__☓

__2 ☓ 2 ☓ 2__☓ 2 ☓ 2

Here, two 2s are left which requires one more 2 to form a triplet. To make 256 a cube, one more 2 is required.

256 ☓ 2 =

__2 ☓ 2 ☓ 2__☓

__2 ☓ 2 ☓ 2__☓

__2 ☓ 2 ☓ 2__= 512 is a perfect cube.

Hence, the smallest natural number by which 256 should be multiplied to make it a perfect cube is 2.

**(iii)**72 =

__2 ☓ 2 ☓ 2__☓ 3 ☓ 3

Here, two 3s are left which are not in a triplet. To make 72 a cube, one more 3 is required.

72 ☓ 3 =

__2 ☓ 2 ☓ 2__☓

__3 ☓ 3 ☓ 3__= 216 is a perfect cube.

Hence, the smallest natural number by which 72 should be multiplied to make it a perfect cube is 3.

**(iv)**675 =

__3 ☓ 3 ☓ 3__☓ 5 ☓ 5

Here, two 5s are left which require one more 5 to forma a triplet.

∴ 675 ☓ 5 =

__3 ☓ 3 ☓ 3__☓

__5 ☓ 5 ☓ 5__= 3375 is a perfect cube.

Hence, the smallest natural number by which 675 should be multiplied to make it a perfect cube is 5.

**(v)**100 = 2 ☓ 2 ☓ 5 ☓ 5

Here, two 2s and two 5s are left which are do not form a triplet. To make 100 a cube, we need one more 2 and one more 5.

∴ 100 ☓ 2 ☓ 5 =

__2 ☓ 2 ☓ 2__☓

__5 ☓ 5 ☓ 5__= 1000 is a perfect cube

Hence, the smallest natural number by which 100 should be multiplied to make it a perfect cube is

2 ☓5 = 10.

**Q3: 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

(ii) 128

(iii) 135

(iv) 192

(v) 704

Answer:

**(i)**81 =

__3 ☓ 3 ☓ 3__☓ 3

Here, one 3 is left which does not form a triplet.

If we divide 81 by 3, then it will become a perfect cube.

∴ 81 ÷ 3 = 27 =

__3 ☓ 3 ☓ 3__is a perfect cube.

Hence, the smallest number by which 81 should be divided to make it a perfect cube is 3.

**(ii)**128 =

__2 ☓ 2 ☓ 2__☓

__2 ☓ 2 ☓ 2__☓ 2

Here, one 2 is left which is not in a triplet.

If we divide 128 by 2, then it will become a perfect cube.

∴ 128 ÷ 2 = 64 =

__2 ☓ 2 ☓ 2__☓

__2 ☓ 2 ☓ 2__is a perfect cube.

Hence, the smallest number by which 128 should be divided to make it a perfect cube is 2.

**(iii)**135 =

__3 ☓ 3 ☓ 3__☓ 5

Here, one 5 is left which is not part of a triplet.

If we divide 135 by 5, then it will become a perfect cube.

Thus, 135 ÷ 5 = 27 =

__3 ☓ 3 ☓ 3__is a perfect cube.

Hence, the smallest number by which 135 should be divided to make it a perfect cube is 5.

**(iv)**192 =

__2 ☓ 2 ☓ 2__☓

__2 ☓ 2 ☓ 2__☓ 3

Here, one 3 is left which is not part of a triplet.

By dividing 192 by 3, it becomes a perfect cube.

Thus, 192 ÷ 3 = 64 =

__2 ☓ 2 ☓ 2__☓

__2 ☓ 2 ☓ 2__is a perfect cube.

Hence, the smallest number by which 192 should be divided to make it a perfect cube is 3.

**(v)**704 =

__2 ☓ 2 ☓ 2__☓

__2 ☓ 2 ☓ 2__☓ 11

Here, one 11 is left which does not form a triplet.

If we divide 704 by 11, then it makes a perfect cube.

Thus, 704 ÷ 11 = 64 =

__2 ☓ 2 ☓ 2__☓

__2 ☓ 2 ☓ 2__is a perfect cube.

Hence, the smallest number by which 704 should be divided to make it a perfect cube is 11.

**Q4: 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?**

Answer:

Volume of the cube of sides 5 cm, 2 cm, 5 cm = 5 cm ☓ 2 cm ☓ 5 cm = (5 ☓ 5 ☓ 2) cm

^{3}

Here, two 5s and one 2 are left which do not form a triplet.

To make it a pefect cube, multiply this expression by 2 ☓ 2 ☓ 5 = 20, then it will become a perfect cube.

i.e. (5 ☓ 5 ☓ 2 ☓ 2 ☓ 2 ☓ 5) = (

__5 ☓ 5 ☓ 5__☓

__2 ☓ 2 ☓ 2__) = 1000 is a perfect cube.

Therefore 20 cuboids of 5 cm, 2 cm, 5 cm are required to form a cube.

Go to babajikenotes.com for notes of jee main or +1 or +2

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