Given,

a_n = n³ – 6n² + 11n – 6, n ∈ N

We can find first three terms of sequence by putting the values of n form 1 to 3.

When n = 1:

a₁ = (1)³ – 6(1)² + 11(1) – 6

⇒ a₁ = 1 – 6 + 11 – 6

⇒ a₁ = 12 – 12

⇒ a₁ = 0

When n = 2:

a₂ = (2)³ – 6(2)² + 11(2) – 6

⇒ a₂ = 8 – 6(4) + 22 – 6

⇒ a₂ = 8 – 24 + 22 – 6

⇒ a₂ = 30 – 30

⇒ a₂ = 0

When n = 3:

a₃ = (3)³ – 6(3)² + 11(3) – 6

⇒ a₃ = 27 – 6(9) + 33 – 6

⇒ a₃ = 27 – 54 + 33 – 6

⇒ a₃ = 60 – 60

⇒ a₃ = 0

This shows that the first three terms of the sequence is zero.

When n = n:

a_n = n³ – 6n² + 11n – 6

⇒ a_n = n³ – 6n² + 11n – 6 – n + n – 2 + 2

⇒ a_n = n³ – 6n² + 12n – 8 – n + 2

⇒ a_n = (n)³ – 3×2n(n – 2) – (2)³ – n + 2

{(a – b)³ = (a)³ – (b)³ – 3ab(a – b)}

⇒ a_n = (n – 2)³ – (n – 2)

Here, n – 2 will always be positive for n > 3

∴ a_n is always positive for n > 3