Answer
$Q(x)$ is continuous on its domain $(-\infty,\sqrt[3]2)\cup(\sqrt[3]2,\infty)$
Work Step by Step
$$Q(x)=\frac{\sqrt[3]{x-2}}{x^3-2}$$
1) Consider the numerator:
The domain of $\sqrt[3]{x-2}$ is $(-\infty,\infty)$
According to Theorem 7, $\sqrt[3]{x-2}$ is continuous on $(-\infty,\infty)$
2) Consider the denominator:
The domain of $x^3-2$ is $(-\infty,\infty)$
According to Theorem 7, $x^3-2$ is continuous on $(-\infty,\infty)$
Overall, both $\sqrt[3]{x-2}$ and $x^3-2$ are continuous on $(-\infty,\infty)$
3) According to part 5 of Theorem 4,
$Q(x)$ is continuous on $(-\infty,\infty)$ except where $(x^3-2)=0$
We see that
$(x^3-2)=0$
$\Leftrightarrow x^3=2$
$\Leftrightarrow x=\sqrt[3]2$
Therefore, $Q(x)$ is continuous on its domain $(-\infty,\sqrt[3]2)\cup(\sqrt[3]2,\infty)$
