Answer
$f \circ g=\sqrt{x^3-2}$
$g \circ f=x^{3/2}-2$
$f \circ f=x^{1/4}$
$g \circ g=x^9-6x^6+12x^3-10$
Work Step by Step
We have: $f(x)=\sqrt{x}$ and $g(x)=x^3-2$.
$f \circ g=f(g(x))=\sqrt{x^3-2}$
$g \circ f=g(f(x))=(\sqrt{x})^3-2=x^{3/2}-2$
$f \circ f=f(f(x))=\sqrt{\sqrt{x}}=x^{1/4}$
$g \circ g=g(g(x))=(x^3-2)^3-2=x^9-6x^6+12x^3-10$
