We have f(x) = 1 – x – x3
⇒ f'(x) = –1 – 3x2 < 0, " x ∈ R
f(x) is strictly decreasing
Now, f( f(x)) = 1 – f(x) – ( f(x))3
⇒ f(f(x)) > f(1 – 5x)
⇒ f(x) < (1 – 5x)
⇒ 1 – x – x3 < (1 – 5x)
⇒ x3 – 4x > 0
⇒ x(x2 – 4) > 0
⇒ x(x + 2)(x – 2) > 0
⇒ x ∈(–2, 0) ∪ (2, ∞)