- Hardy - Ramanujan Number
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- Cube Numbers
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- Prime Factorization of Cubes
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- Smallest multiple that is a perfect cube
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- Cube Root
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- How to Find Cube root

In this page we will explain the topics for the chapter 7 of Cube and Cube Roots Class 8 Maths.We have given quality notes and video to explain various things so that students can benefits from it and learn maths in a fun and easy manner, Hope you like them and do not forget to like , social share
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expressed as sum of two cubes in two different ways.

1729 = 1728 + 1 = 12

1729 = 1000 + 729 = 10

1729 is the smallest Hardy– Ramanujan Number. There are an infinitely many such numbers. Few are 4104 (2, 16; 9, 15), 13832 (18, 20; 2, 24), Check it with the numbers given in the brackets

Example

1=1^{3}

8=2^{3}

27=3^{3}

Or if a natural number 1=1

8=2

27=3

The numbers 1, 8, 27, 125 ... are cube numbers. These numbers are also called

8= 2×2×2 (Triplet of prime factor 2)

216 = (2 × 2 × 2) × (3 × 3 × 3) ( Triplet of 2 and 3)

Each prime factor of a number appears three times in the prime factorization of its cube.

Find the smallest multiple that will make 392 perfect cube

392 = 2 × 2 × 2 × 7 × 7

The prime factor 7 does not appear in a group of three. Therefore, 392 is not a perfect ube. To make its a cube, we need one more 7. In that case

392 × 7 = 2 × 2 × 2 × 7 × 7 × 7 = 2744 which is a perfect cube

So we know that

27=3

Cube root of 27

Cube root is denoted by expression

We will get same prime number occurring in pair for perfect square number. Square root will be given by multiplication of prime factor occurring in pair

Consider

1331

1331= (11×11×11)

5832

5832= (2 × 2 × 2) × (3 × 3× 3) × (3 × 3× 3)

The given number is 17576.

17 576. In this case one group i.e., 576 has three digits whereas 17 has only two digits.

The digit 6 is at its one’s place.

We take the one’s place of the required cube root as 6.

Cube of 2 is 8 and cube of 3 is 27. 17 lies between 8 and 27.

The smaller number among 2 and 3 is 2.

The one’s place of 2 is 2 itself. Take 2 as ten’s place of the cube root of

17576.

Thus,

1) for any Positive integer m, m

2) For any negative integer m, m

3) Cubes can be written as Addition consecutive odd numbers

1 = 1 = 1

3 + 5 = 8 = 2

7 + 9 + 11 = 27 = 3

13 + 15 + 17 + 19 = 64 = 4

21 + 23 + 25 + 27 + 29 = 125 = 5

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