Let $f(x)=1-x-x^3$, then the real values of x satisfying the inequality $1-f(x)-f^3(x)>f(1-5 x)$, is

$(0,2)$  $(-2,2)$ $(-2,1) \cup(1, \infty)$ $(-\infty,-2) \cup(0,2)$

$(-\infty,-2) \cup(0,2)$

We have, $f(x)=1-x-x^3$  $f'(x)=-1-3 x^2<0$ (decreasing function) so $1-f(x)-f^3(x)>f(1-5 x)$ $=f(f(x))>f(1-5 x)$  $1-x-x^3<1-5 x$  [as $f(x) < 1-5 x$] so $x^3+x>5x$ $x^3-4x>0$ $x(x^2-4)>0$ so for this condition $x∈(-2, 0) \cup(2, ∞)$