Answer
$ \left (\frac{f}{g} \right )(x)=2x\sqrt [3]{2x}$.
$(-\infty,0)\cup(0,\infty)$.
Work Step by Step
The given functions are
$f(x)=\sqrt[3] {32x^6}$ and $g(x)=\sqrt [3]{2x^2}$
$\Rightarrow \left (\frac{f}{g} \right )(x)=\frac{f(x)}{g(x)}$
Substitute both functions.
$\Rightarrow \left (\frac{f}{g} \right )(x)=\frac{\sqrt[3] {32x^6}}{\sqrt [3]{2x^2}}$
Divide the radicands and retain the common index.
$\Rightarrow \left (\frac{f}{g} \right )(x)=\sqrt[3] {\frac{32x^6}{2x^2}}$
Divide factors in the radicand. Subtract exponents on common bases.
$\Rightarrow \left (\frac{f}{g} \right )(x)=\sqrt[3] {16x^{6-2}}$
Simplify.
$\Rightarrow \left (\frac{f}{g} \right )(x)=\sqrt[3] {16x^{4}}$
The domain is all positive and negative real numbers excluding zero.
The interval notation is $(-\infty,0)\cup(0,\infty)$.
$\Rightarrow \left (\frac{f}{g} \right )(x)=2x\sqrt [3]{2x}$.
