# NCERT Class 8 Mathematics Solutions: Chapter 7 – Cubes and Cube Roots Exercise 7.1 Part 2 (For CBSE, ICSE, IAS, NET, NRA 2022)

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**Question: 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) 675

(v) 100

**Answer**:

**(i) 243**

3 | 243 |

3 | 81 |

3 | 27 |

3 | 9 |

3 | 3 |

1 |

Prime factors of

To make 243 a cube, one more 3 is required.

Hence, the number 243 multiply by 3 to obtain a perfect cube.

**(ii) 256**

2 | 256 |

2 | 128 |

2 | 64 |

2 | 32 |

2 | 16 |

2 | 8 |

2 | 4 |

2 | 2 |

1 |

Prime factors of 256 =

To make 256 a cube, one more 2 is required.

Hence, the number 256 multiply by 2 to obtain a perfect cube.

**(iii) 72**

2 | 72 |

2 | 36 |

2 | 18 |

3 | 9 |

3 | 3 |

1 |

Prime factors of

To make 72 a cube, one more 3 is required.

Hence, the number 72 multiply by 3 to obtain a perfect cube.

**(iv) 675**

3 | 675 |

3 | 225 |

3 | 75 |

5 | 25 |

5 | 5 |

1 |

Prime factors of

To make 675 a cube, one more 5 is required.

Hence, the number 675 multiply by 5 to obtain a perfect cube.

**(v) 100**

2 | 100 |

2 | 50 |

5 | 25 |

5 | 5 |

1 |

Prime factors of

To make 100 a cube, we require one more 2 and one more 5.

Hence, the number 100 multiply by to obtain a perfect cube.