By the mathematical induction, we have to prove that n³ - n is divisible by 6

Given that S(n) = n³ - n divisible by 6.

Let n =1 then we get 0

which is divisible by 6

So, S(1) is true.

Let us assume that n = k

S(k) = k³ - k

which is divisible by 6.

So, S(k) is true.

=> (k³ - k) / 6 = m (integer)

=> (k³ - k) = 6m

=> k³ = 6m + k  .............1

now we have to prove that n = k+1

⇒ (k + 1)³ - (k + 1)

⇒ (k³ + 3k² + 3k + 1) - (k + 1)

subsitute equation 1 in above equation then

=> 6m + k + 3k² + 2k

=> 6m + 3k² + k

=> 6m + 3k(k + 1) (Since k(k + 1) = 2p is an even number p is natural number)

=> 6m +3 * 2p

=> 6(m + p)

which is divisible by 6

So, S(k + 1) is true.

Hence, By the mathematical induction, it is true that n³ - n is divisible by 6 for

n∈N.