Why can I simplify radicals? (eg, $\sqrt{153} = \sqrt{3}\cdot \sqrt{3}\cdot \sqrt{17}$) I know it might sound like a ridiculously easy question to answer, but I just can't put two and two together for some reason. 
Say for example you have:
$$\sqrt{153}$$
You can break it down to 
$$\sqrt{3}\cdot \sqrt{3}\cdot \sqrt{17}$$ 
(and of course you can further simplify it, but we'll just leave it at that for now). 
Why can I do that?  

Ignoring some (ahem) complexities, we can say, for non-negative(!) $x$,

$\sqrt{x}\;$ is the non-negative(!) number that multiplies by itself to give  $\;x$.

With that in mind, consider what happens when we multiply $\sqrt{x}\cdot\sqrt{y}$ by itself:
$$\begin{align}
\left(\;\color{red}{\sqrt{x}}\cdot\color{blue}{\sqrt{y}}\;\right)\cdot\left(\;\color{red}{\sqrt{x}}\cdot\color{blue}{\sqrt{y}}\;\right) &=
\color{red}{\sqrt{x}}\cdot\color{blue}{\sqrt{y}}\cdot\color{red}{\sqrt{x}}\cdot\color{blue}{\sqrt{y}} \\
&= \color{red}{\sqrt{x}}\cdot\color{red}{\sqrt{x}}\cdot\color{blue}{\sqrt{y}}\cdot\color{blue}{\sqrt{y}} \\
&= \left(\;\color{red}{\sqrt{x}}\cdot\color{red}{\sqrt{x}}\;\right)\cdot\left(\;\color{blue}{\sqrt{y}}\cdot\color{blue} {\sqrt{y}}\;\right) \\
&= \color{red}{x}\cdot\color{blue}{y}
\end{align}$$
Since $\sqrt{x}$ and $\sqrt{y}$ are non-negative, so is their product. Therefore, we have shown that

$\sqrt{x}\cdot\sqrt{y}\;$ is the non-negative(!) number that multiplies by itself to give $\;x\cdot y$.

More simply,

$$\sqrt{x\cdot y} = \sqrt{x}\cdot\sqrt{y}$$

The idea extends to show that
$$\sqrt{x\cdot y\cdot z} = \sqrt{x}\cdot\sqrt{y}\cdot\sqrt{z}$$
and so forth. (By a completely-analogous argument, this kind of thing works for cube-roots, and fourth-roots, and $n$-th roots in general.) This is precisely what allows you to simplify radicals.
