# Exercise 6.2 class 8 solution

#### Exercise 6.2 class 8 solution-Introduction

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#### Exercise 6.2 class 8 solution-Some interesting patterns

(a) Adding consecutive odd numbers :

1=1=1^3

3+5=8=2^3

7+9+11=27=3^3

13+15+17+ 19=64=4^4

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

31+33 +35+37+39 +41 = 216 =6^6

There are patterns in the sums of consecutive odd cubes. For instance, the sum of the first n odd cubes is always a perfect square:

(b) Cubes and their Prime factors :

Consider the following Prime factorization of some numbers and their cubes Prime factorization of a number Prime factorization of its cube

Observe that each prime factor of a number appears three times in the prime factorization of its cube.

In the prime factorization of any number, if each factor appears three times, then

the number is perfect cube.

#### Exercise 6.2 class 8 solution- what is prime factorization method ?

Prime factorization is a fundamental concept in number theory and has various applications in mathematics and science, including simplifying fractions, finding the greatest common divisor (GCD), and solving certain mathematical problems.

To find the prime factorization of 512 using the prime factorization method, follow these steps:

1. Start by dividing 512 by the smallest prime number, which is 2, and continue dividing by 2 until you can no longer divide evenly:
2. 512 ÷ 2 = 256
3. 256 ÷ 2 = 128
4. 128 ÷ 2 = 64
5. 64 ÷ 2 = 32
6. 32 ÷ 2 = 16
7. 16 ÷ 2 = 8
8. 8 ÷ 2 = 4
9. 4 ÷ 2 = 2
10. 2 ÷ 2 = 1
1. Now, express 512 as a product of its prime factors:512 = 2^9

So, the prime factorization of 512 is 2^9. This means that 512 can be represented as the product of 2 raised to the power of 9, making it a perfect cube because 9 is a multiple of 3. In this case, it’s 2^3 raised to the power of 3, which is indeed a perfect cube.

#### Exercise 6.2 class 8 solution- exercise preview

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#### Exercise 6.2 class 8 solution- solution pdf

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6.2

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