CBSE Class 10 Answered

Any positive integer is of the form 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4, 6m + 5 for some positive integer n.

When n = 6m,

n³ - n = (6m)³ - 6m = 216 m³ - 6m

= 6m(36m² - 1)

= 6q, where q = m(36m² -1)

n³ - n is divisible by 6

When n = 6m + 1,

n³ - n = n(n² - 1) = n (n - 1) (n + 1)

= (6m + 1) (6m) (6m + 2)

= 6m(6m + 1) (6m + 2)

= 6q, where q = m(6m + 1) (6m + 2)

n³ - n is divisible by 6

When n = 6m + 2,

n³ - n = n (n - 1) (n + 1)

= (6m + 2) (6m + 1) (6m + 3)

= (6m + 1) (36 m² + 30m + 6)

= 6m (36 m² + 30m + 6) + 1 (36m² + 30m + 6)

= 6[m (36m² + 30m + 6)] + 6 (6m² + 5m + 1)

= 6p + 6q,

where p = m (36m² + 30m + 6)

q = 6m² + 5m + 1

n³ - n is divisible by 6

When n = 6m + 3

n³ - n = (6m + 3)³ - (6m + 3)

= (6m + 3) [(6m + 3)² - 1]

= 6m [6m + 3)² - 1] + 3 [(6m + 3)² - 1]

= 6 [m [(6m + 3)² - 1] + 3 [36m² + 36m + 8]

= 6 [m [(6m + 3)² - 1] + 6 [18m² + 18m + 4]

= 6p + 3q,

where p = m[(6m + 3)² - 1]

q = 18m² + 18m + 4

n³ - n is divisible by 6

When n = 6m + 4

n³ - n = (6m + 4)³ - (6m + 4)

= (6m + 4) [(6m + 4)² - 1]

= 6m [(6m + 4)² - 1] + 4 [(6m + 4)² - 1]

= 6m [(6m + 4)² - 1] + 4 [36m² + 48m + 16 - 1]

= 6m [(6m + 4)² - 1] + 12 [12m² + 16m + 5]

= 6p + 6q,

where p = m [(6m + 4)² - 1]

q = 2 (12 m² + 16m + 5)

n³ - n is divisible by 6

When n = 6m + 5

n³ - n = (6m + 5) [(6m + 5)² - 1]

= 6m [(6m + 5)² - 1] + 5 [(6m + 5)² - 1]

= 6m [(6m + 5)² - 1] + 5 [36m² + 60m + 24]

= 6p + 30q

= 6 (p + 5q),

where p = m [(6m + 5)² - 1]

q = 6m² + 10m + 4

n³ - n is divisible by 6

Hence, n³ - n is divisible by 6, for any +ve integer n.