Answer
With $g(x)=1-x^{3}$, and $f(x)=|x|$,
$H(x)=f[g(x)]=f\circ g(x)$
Work Step by Step
If we were to calculate by hand,
we would subtract the cube of the number from 1$\qquad$...$ g(x)=1-x^{3}$,
then find its absolute value$\qquad$ .... $f(x)=|x|$
In this way, $H(x)=f(g(x))=f\circ g(x)$