Simplify $\sqrt[4]{\frac{162x^6}{16x^4}}$ is $\frac{3\sqrt[4]{2x^2}}{2}$ (I've been posting a lot today and yesterday, not sure if too many posts are frowned upon or not. I am studying and making sincere efforts to solve on my own and only post here as a last resort)
I'm asked to simplify $\sqrt[4]{\frac{162x^6}{16x^4}}$ and am provided the text book solution $\frac{3\sqrt[4]{2x^2}}{2}$.
I arrived at $\frac{3\sqrt[4]{2x^6}}{2x^4}$. I cannot tell if this is right and that the provided solution is just a further simplification of where I've gotten to, or if I'm off track entirely.
Here is my working:
$\sqrt[4]{\frac{162x^6}{16x^4}}$ = $\frac{\sqrt[4]{162x^6}}{\sqrt[4]{16x^4}}$
Denominator: $\sqrt[4]{16x^4}$ I think can be simplified to $2x^4$ since $2^4$ = 16
Numerator: $\sqrt[4]{162x^6}$ I was able to simplify (or over complicate) to $3\sqrt[4]{2}\sqrt[4]{x^6}$ since:
$\sqrt[4]{162x^6}$ = $\sqrt[4]{81}$ * $\sqrt[4]{2}$ * $\sqrt[4]{x^6}$ = $3 * \sqrt[4]{2} * \sqrt[4]{x^6}$
Thus I got:
$\frac{3\sqrt[4]{2}\sqrt[4]{x^6}}{2x^4}$ which I think is equal to $\frac{3\sqrt[4]{2x^6}}{2x^4}$ (product of the radicals in the numerator).
How ca I arrive at the provided solution $\frac{3\sqrt[4]{2x^2}}{2}$?

$$\sqrt[4]{\frac{162x^6}{16x^4}}=\sqrt[4]{\frac{81\cdot2x^2}{16}}=\frac{3\sqrt[4]{2x^2}}{2}.$$
