Math Help - Simplifying root denominators

1. Simplifying root denominators

This expression I understand perfectly

$\frac{x}{\sqrt[4]x}=\frac{x}{\sqrt[4]x}*\frac{\sqrt[4]x^3}{\sqrt[4]x^3}$

= ${\sqrt[4]x^3}$

This however I am completely lost...

$\frac{9}{\sqrt[5]8}=\frac{9}{\sqrt[5]2^3}=\frac{9}{\sqrt[5]2^3}*\frac{\sqrt[5]2^2}{\sqrt[5]2^2}=\frac{9\sqrt[5]4}{\sqrt[5]2^5}=\frac{9\sqrt[5]4}{2}$

Why must the $\sqrt[5]8$ be broken down into $\sqrt[5]2^3$, and how are you supposed to know, and how is this done?

I would solve it like this, why wouldnt this work?
$\frac{9}{\sqrt[5]8}=\frac{9}{\sqrt[5]8}*\frac{\sqrt[5]8^4}{\sqrt[5]8^4}=\frac{9\sqrt[5]4096}{\sqrt[5]8^4}=\frac{9\sqrt[5]4}{2}
$

2. Originally Posted by allyourbass2212
This expression I understand perfectly

$\frac{x}{\sqrt[4]x}=\frac{x}{\sqrt[4]x}*\frac{\sqrt[4]x^3}{\sqrt[4]x^3}$

= ${\sqrt[4]x^3}$

This however I am completely lost...

$\frac{9}{\sqrt[5]8}=\frac{9}{\sqrt[5]2^3}=\frac{9}{\sqrt[5]2^3}*\frac{\sqrt[5]2^2}{\sqrt[5]2^2}=\frac{9\sqrt[5]4}{\sqrt[5]2^5}=\frac{9\sqrt[5]4}{2}$

Why must the $\sqrt[5]8$ be broken down into $\sqrt[5]2^3$, and how are you supposed to know, and how is this done?

I would solve it like this, why wouldnt this work?
$\frac{9}{\sqrt[5]8}=\frac{9}{\sqrt[5]8}*\frac{\sqrt[5]8^4}{\sqrt[5]8^4}=\frac{9\sqrt[5]4096}{\sqrt[5]8^4}=\frac{9\sqrt[5]4}{2}
$
$\frac{9}{\sqrt[5]{8}} = \frac{9}{\sqrt[5]{8}} \times \frac{\sqrt[5]{8^4}}{\sqrt[5]{8^4}}$

$\frac{9\sqrt[5]{8^4}}{\sqrt[5]{8^5}}$ $=\frac{9(\sqrt[5]{8^4})}{8}=\frac{9(\sqrt[5]{2^{12}})}{8}$

since $\sqrt[5]{8^4}=\sqrt[5]{8*8*8*8}=\sqrt[5]{2^3(2^3)(2^3)(2^3)}=\sqrt[5]{(2^3)^4}=$ $\sqrt[5]{2^{3(4)}}=\sqrt[5]{2^{12}}$

$\frac{9(\sqrt[5]{2^{12}})}{8}=\frac{9\sqrt[5]{2^{10}(2^2)}}{8}=\frac{9(4)\sqrt[5]{4}}{8}=\frac{9\sqrt[5]{4}}{2}$

he broke

$\sqrt[5]{8} = \sqrt[5]{2^3}$ to make it easier so you will multiply with $\frac{\sqrt[5]{2^2}}{\sqrt[5]{2^2}}$

but you see if we do not broke it it will be more harder ...