College Algebra, Chapter 3, 3.6, Section 3.6, Problem 56

Given $F(x) = \sqrt[3]{\sqrt{x}-1}$, find functions $f,g$ and $h$ such that $F = f \circ g \circ h$

Since the formula for $F$ says to first take the square root. Then, subtract 1 and lastly take the cube root. So we let,
$h(x) = \sqrt{x}, \quad g(x) = x- 1, \quad$ and $f(x) = \sqrt[3]{x}$

$
\begin{equation}
\begin{aligned}
\text{Then } (f\circ g\circ h)(x) &= f(g(h(x))) && \text{Definition of } f \circ g \circ h\\
\\
(f\circ g\circ h)(x) &= f(g(\sqrt{x})) && \text{Definition of } h\\
\\
(f\circ g\circ h)(x) &= f(\sqrt{x}-1) && \text{Definition of } g\\
\\
(f\circ g\circ h)(x) &= \sqrt[3]{\sqrt{x}-1} && \text{Definition of } f\\
\\
(f\circ g\circ h)(x) &= F(x)
\end{aligned}
\end{equation}
$