Part I :

Square term is always positive and (x + 1)² ≠ 0. So, x > 0 and x ≠ -1

Part II :

Square term is always positive and (x – 1)² ≠ 0. So, x < 0 and x ≠ 1

So, by taking union of part I and part II we have final solution i.e., x ∈ R – {-1, 0, 1}