# NCERT Class 8 Mathematics Solutions: Chapter 7 – Cubes and Cube Roots Exercise 7.1 Part-3 (For CBSE, ICSE, IAS, NET, NRA 2022)

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**Question: 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

**Answer**:

**(i) 81**

3 | 81 |

3 | 27 |

3 | 9 |

3 | 3 |

1 |

Prime Factors of

Here, one 3 is left which is not in a triplet.

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

Thus, is a perfect cube.

Hence, 81 divided by **3** to make it a perfect cube.

**(ii) 128**

2 | 128 |

2 | 64 |

2 | 32 |

2 | 16 |

2 | 8 |

2 | 4 |

2 | 2 |

1 |

Prime Factors of

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

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

Thus, is a perfect cube.

Hence, 128 divided by **2** to make it a perfect cube.

**(iii) 135**

3 | 135 |

3 | 45 |

3 | 15 |

5 | 5 |

1 |

Prime Factors of

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

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

Thus, is a perfect cube.

Hence, 135 divided by **5** to make it a perfect cube.

**(iv) 192**

2 | 192 |

2 | 96 |

2 | 48 |

2 | 24 |

2 | 12 |

2 | 6 |

3 | 3 |

1 |

Prime Factors of

Here, one 3 is left which is not in a triplet.

If we divide 192 by 3, then it will become a perfect cube

Thus, is a perfect cube.

Hence, 192 divided by **3** to make it a perfect cube.

**Question: 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?

**Answer**:

Given numbers

Since, factors of 5 and 2 both are not in group of three.

Therefore, the number must be multiplied by to make it a perfect cube. Hence he needs **20** cuboids.