<s>[INST] Rationalize the denominator: $(\frac{3}{2x^2}) ^{1/4}$ My answer is $\frac{(6x^2)^{1/4}}{ 2x^2}$ 
However the book's answer is $\frac{(24x^2)^{1/4}}{2x}$
Where did the book get that from? 

Here's my work: 
$$\sqrt[4]{\frac{3}{2x^2}}= \frac{\sqrt[4]{3}}{\sqrt[4]{2x^2}}\cdot \frac{\sqrt[4]{2x^2}}{\sqrt[4]{2x^2}}$$
$$= \frac{\sqrt[4]{6x^2}}{2x^2}$$
Since the product of two roots with the same index is the root of the product:
$$\sqrt[N]{a} \cdot \sqrt[N]{b} = \sqrt[N]{ab}$$
 [/INST] The simplest way is to multiply numerator and denominator by $(2x^2)^{\color{red}3/4}$, just because $\frac34+\frac14=1$:
\begin{align*}
\biggl(\frac3{2x^2}\biggr)^{1/4}&=\frac{3^{1/4}}{(2x^2)^{1/4}}=\frac{3^{1/4}(2x^2)^{3/4}}{2x^2}=\frac{(3\cdot8x^6)^{1/4}}{2x^2}\\
&=\frac{(24x^2)^{1/4}\cdot x}{2x^2}=\frac{\Bigl(\cfrac32x^2\Bigr)^{1/4}}x.
\end{align*}
</s>