Let f(x)=1-x-x^3, then the real values of x satisfying the inequality…

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

$x∈(-2, 0) \cup(2, ∞)$