<s>[INST] 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 $\frac{3\sqrt[4]{2x^2}}{2}$?
 [/INST] You made an error:
$$\sqrt[4]{16x^4} = 2\vert x\vert$$
because $\sqrt[4]{16x^4} = \sqrt[4]{(2x)^4}$. (Note the absolute value sign since the value returned is positive regardless of whether $x$ itself is positive or negative.) The rest is fine, so from here, you get
$$\frac{3\sqrt[4]{2x^6}}{2\vert x\vert} = \frac{3\sqrt[4]{2x^4x^2}}{2x} =  \frac{3\vert x\vert\sqrt[4]{2x^2}}{2\vert x\vert} = \frac{3\sqrt[4]{2x^2}}{2}$$
As shown in the other answer, it is usually better to simplify within the radical so you don’t mess up with absolute values (for even indices).
$$\sqrt[4]{\frac{162x^6}{16x^4}} = \sqrt[4]{\frac{2\cdot3^4x^2}{2^4}} = \frac{3\sqrt[4]{2x^2}}{2}$$
</s>