The document Points to Remember- Cubes and Cube Roots Class 8 Notes | EduRev is a part of the Class 8 Course Mathematics (Maths) Class 8.

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**Facts That Matter**

- If we multiply a number by itself three times, the product so obtained is called the perfect cube of that number.
- There are only 10 perfect cubes from 1 to 1000.
- Cubes of even numbers are even and those of odd numbers are odd.
- The cube of a negative number is always negative.
- If the digit in the one’s place of a number is 0, 1, 4, 5, 6 or 9, then its cube will end in the same digit.
- If the digit in the one’s place of a number is 2, then the ending digit of its cube will be 8 and vice-versa.
- If the digit in the one’s place of a number is 3, then the ending digit of its cube will be 7 and vice-versa.
- If the prime factors of a number cannot be made into groups of 3, it is not a perfect cube.
- The symbol ∛ denotes the cube root of a number, such as ∛64 = 4.

**We Know That**

The square of a number is obtained by multiplying the number by itself, i.e.

2 * 2 = 2^{2}

4 * 4 = 4^{2}

5 * 5 = 5^{2}

Finding square root is the inverse operation of finding the square of a number, i.e.

A natural number multiplied by itself three times gives cube of that number, e.g.

1 * 1 * 1 = 1

2 * 2 * 2 = 8

3 * 3 * 3 = 27

4 * 4 * 4 = 64

The numbers 1, 8, 27, 64, … are called cube numbers or perfect cubes.**Properties of Perfect Cubes****Property I:** If the digit in the one’s place of a number is 0, 1, 4, 5, 6 or 9, then the digit in the one’s place of its cube will also be the same digit.**Property II:** If the digit in the one’s place of a number is 2, the digit in the one’s place of its cube is 8, and vice-versa.**Property III: **If the digit in the one’s place of a number is 3, the digit in the one’s place of its cube is 7 and vice-versa.**Property IV: **Cubes of even natural numbers are even.**Examples:**

Number | Cube | Number | Cube |

8 36 | 512 46656 | 12 18 | 1728 5832 |

**Property V: **Cubes of odd natural numbers are odd.**Examples:**

Number | Cube | Number | Cube |

9 35 | 729 42875 | 13 17 | 2197 4913 |

**Property VI: **Cubes of negative integers are negative.**Examples:**

Number | Cube | Number | Cube |

–51 –14 | –132651 –2744 | –11 –22 | –1331 –10648 |

**Patterns in Cubes****I . ****Adding consecutive odd numbers:**

Number | Cube | Sum of consecutive odd numbers |

1 | 1 ^{3} = 1 | 1 |

2 | 2 ^{3 }= 8 | 3 + 5 |

3 | 3 ^{3} = 27 | 7 + 9 + 11 |

4 | 4 ^{3} = 64 | 13 + 15 + 17 + 19 |

5 | 5 ^{3} = 125 | 21 + 23 + 25 + 27 + 29 |

Note that we start with [n * (n – 1) + 1] odd number.**II.** **Difference of two consecutive cubes:**

2^{3} – 1^{3} = 1 + 2 * 1 * 3

3^{3} – 2^{3} = 1 + 3 * 2 * 3

4^{3} – 3^{3} = 1 + 4 * 3 * 3

5^{3} – 4^{3} = 1 + 5 * 4 * 3

**III.** **Cubes and their prime factor:**

[Each prime factor of the number appear three times in its cube.]

Number | Prime factorisation | Cube | Prime factors of the cube in its prime factorisations |

4 | 2 * 2 | 4 | 2 * 2 * 2 * 2 * 2 * 2 = 2 |

12 | 2 * 2 * 3 | 12 | 2 * 2 * 2 * 2 * 2 * 2 * 3 * 3 * 3 = 2 |

15 | 3 * 5 | 15 | 3 * 3 * 3 * 5 * 5 * 5 = 3 |

18 | 2 * 3 * 3 | 18 | 2* 2 * 2 * 3 * 3 * 3 * 3 * 3 * 3 = 2 |

21 | 3 * 7 | 21 | 3 * 3 * 3 * 7 * 7 * 7 = 3 |

**Example 1:** Is 500 a perfect cube?**Solution:** 500 = 5 * 5 * 5 * 2 * 2

∵ In the above prime factorisation 2 * 2 remain after grouping the prime factors in triples.

∴ 500 is not a perfect cube.

**Example 2:** Is 1372 a perfect cube? If not, find the smallest natural number by which 1372 must be multiplied so that the product is a perfect cube.**Solution:** We have 1372 = 2 * 2 * 7 * 7 * 7

Since, the prime factor 2 does not appear in a group of triples.

∴ 1372 is not a perfect cube.

Obviously, to make it a perfect cube we need one more 2 as its factor.

i.e. [1372] * 2 = [2 * 2 * 7 * 7 * 7] * 2

or

2744 = 2 * 2 * 2 * 7 * 7 * 7

which is a perfect cube.

Thus, the required smallest number = 2.

**Example 3:** Is 31944 a perfect cube? If not then by which smallest natural number should 31944 be divided so that the quotient is a perfect cube?**Solution:** We have 31944 = 2 * 2 * 2 * 3 * 11 * 11 * 11

Since, the prime factors of 31944 do not appear in triples as 3 is left over.

∴ 31944 is not a perfect cube. Obviously, 31944 / 3 will be a perfect cube

i.e. [31944] ÷3 = [2 * 2 * 2 * 3 * 11 * 11 * 11] /3

or

10648 = 2 * 2 * 2 * 11 * 11 * 11

∴ 10648 is a perfect cube.

Thus, the required least number = 3.

**Examples:**

Numbers | Number ending in | Perfect cube | One’s digit of the cube |

31 22 33 24 35 36 27 58 29 10 | 1 2 3 4 5 6 7 8 9 0 | 29791 10648 35937 13824 42875 46656 19683 195112 24389 1000 | 1 8 7 4 5 6 3 2 9 0 |

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