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:
$$\sqrt[N]{a} \cdot \sqrt[N]{b} = \sqrt[N]{ab}$$
 A: Your error is when you go from the first line to the second.  We have
$$\sqrt[4]{\frac{3}{2x^2}}= \frac{\sqrt[4]{3}}{\sqrt[4]{2x^2}}\cdot \frac{\sqrt[4]{2x^2}}{\sqrt[4]{2x^2}} = \frac{\sqrt[4]{6x^2}}{\sqrt[4]{4x^4}}$$
You can re-express the denominator as $\sqrt[4]{4}\sqrt[4]{x^4} = |x|\sqrt[4]{4}$, which is almost free of radicals (see my note below for where the absolute value came from).  Can you take it from here?

Looking at your work, you seem to be using the (erroneous) rule $\sqrt[N]{a} \cdot \sqrt[N]{a} = a$.  This is true for $N=2$, but not otherwise (this problem has $N=4$).

Edit: Technically, $\sqrt[4]{x^4} = x$ only holds if $x \ge 0$; the more general solution is $\sqrt[4]{x^4} = |x|$, since $\sqrt[4]{\cdot} \ge 0$.  You may have been given that $x>0$ for the problem, at which point $|x| = x$ and there is no difference, but it's a good idea to keep this fact in mind.
A: 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*}
