Let P(n) = n³ – 7n + 3

Now, P(n): n³ – 7n + 3 is divisible by 3 for all n ϵ N

Step1:

P(1) = 1 – 7 + 3 = -3

Thus, P(1) is divisible by 3

Step2:

Let, P(m) be divisible by 24

Then, n³ – 7n + 3 = 3λ, where λ ϵ N.

Now, we need to show that P(m+1) is true whenever P(m) is true.

So, P(m+1) = (n+1)³ – 7(n+1) + 3

= n³+3n²+3n+1-7n-7+3

= n³– 7n + 3 +3n²+3n+1-7

= 3λ+3(n²+n-2)

= 3(λ+n²+n-2)

Thus, P(m+1) is true.

So, by the principle of mathematical induction, P(n) is true for all n ϵN.