Answer is : D. (1, ∞)
Given:
⇒ f(x) = (x+1)3.(x - 3)3
⇒ f’(x) = 3(x+1)2(x - 3)3 + 3(x - 3)3 (x+1)3
Put f’(x) = 0
⇒ 3(x+1)2(x - 3)3 = -3(x - 3)2(x+1)3
⇒ x - 3 = -(x+1)
⇒ 2x = 2
⇒ x = 1
When x>1 the function is increasing.
x<1 function is decreasing.
So, f(x) is increasing in (1, ∞).