Answer
continuous on $(-\infty, \sqrt[3]{2})\cup(\sqrt[3]{2}, \infty)$
Work Step by Step
Let $v(x)=x-2$
It is continuous everywhere, by theorem 7
(polynomials, rational functions, root functions, trigonometric functions are are continuous on their domains)
Let $u(x)=\sqrt[3]{x}. $
It is continuous everywhere, by theorem 7.
Let$ f(x)=u(v(x))=\sqrt[3]{x-2}.$
It is continuous everywhere, by theorem 9 (composite function)
Let $g(x)=x^{3}-2$, a polynomial,
continuous everywhere, by theorem 7.
$Q(x)=\displaystyle \frac{f(x)}{g(x)}$ has domain:$ g(x)\neq 0,$
$x\neq\sqrt[3]{2}$
Domain=$(-\infty, \sqrt[3]{2})\cup(\sqrt[3]{2}, \infty)$
It is continuous on its domain by Th.4.5
(If $f$ and $g$ are continuous, then 5. $\displaystyle \frac{f}{g}$ if $g(a)\neq 0$ is continuous on its domain.
