Question

show that n3 - n is divisible by 8 where n is an odd positive integer​

Answer 1

Given: n³ - n

To find: Show that n³ - n is divisible by 8, n is an odd positive integer​?

Solution:

• Lets simplify the given term, we get

            n³ - n

            n x (n² - 1)

            n x (n - 1) x (n + 1)  

• Let the odd positive integer be 2q+1, where q is an integer.

• Now squaring it, according to the term n² - 1,

           (2q+1)² - 1²

           (2q+1-1) x (2q+1+1)

           2q x (2q + 2)

           2q x 4 x (q+1)

• Now, either q is odd or q+1 is even, so in this case q+1 can be written as 2m and multiplying by 4 we get 8mq, which is divisible by 8.

• And the another case is either q is even or q+1 is odd, so in this case q can be written as 2m and after multiplying we get 8m(q+1) which is divisible by 8 again.

Answer:

So, by above steps we can prove that n³ - n is divisible by 8 where n is an odd positive integer​.

Answer 2

n³ - n is divisible by 8 where n is an odd positive integer​:

Given:

n³ - n

Solution:

n³ - n = n(n² - 1)

Let the value of n be any odd positive integer​.

Let's assume that "n = 3"

n(n² - 1) = 3(9 - 1)

∴ n(n² - 1) = 24

Thus, the number '24' is divisible by 8.

On assuming any odd positive integer as the value of n, then the number is divisible by 8.

Hence proved.