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

100 = 2 × 2 × 5 × 5

Here, each prime factor is not appearing as many times as a perfect multiple of 3. Two 2s and two 5s are remaining after grouping the triplets. Therefore, 100 is not a perfect cube.

(ii) 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__

Here, as each prime factor is appearing as many times as a perfect multiple of 3, therefore, 46656 is a perfect cube.