For any positive integer n, prove that n³ - n is divisible by 6

Solution:

Consider a = n³ - n.

a = n - (n² - 1)

a = n - (n - 1)(n + 1)

Using the algebraic identity (a² - b²) = (a - b)(a + b)

a = (n - 1) n (n + 1).

If a number is divisible by both 2 and 3, then it is divisible by 6.

Three consecutive integers are n - 1, n and n + 1.

a = (n - 1)n (n + 1) is a product of three consecutive integers.

One among these should be divisible by 2 and the other one should be divisible by 3.

a is divisible by both 2 and 3.

Therefore, a = n³ - n is divisible by 6

✦ Try This: For any positive integer n, prove that n² - n is divisible by 2

Case 1:

When n = 2q

n² - n = (2q)² - 2q = 4q² - 2q = 2q(2q - 1).

n² - n = 2r where r = q(2q - 1).

Therefore, n² - n is divisible by 2

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 1

NCERT Exemplar Class 10 Maths Exercise 1.4 Problem 4

For any positive integer n, prove that n³ - n is divisible by 6

Summary:

For any positive integer n, n³ - n is divisible by 6. Hence proved

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