The correct option is C −1<x<4
Given, (x1)2(x+1)3(x4)<0
Since square of a number cannot be <0
This means, one of (x+1)3 or (x4) is <0
Now, (x+1)3<0 and x−4>0 => x<−1 and x>4 --- (1)
OR (x+1)3>0 and x−4<0 => x>−1 and x<4 --- (2)
From solutions of both possibilities, −1<x<4