Is there any way to simplify the following expression into a non-radical form? Is there any way to simplify the following expression into a non-radical form?
$$
\sqrt{x_{1}^2-2x_{1}x_{2}+x_{2}^2+y_{1}^2-2y_{1}y_{2}+y_{2}^2+z_{1}^2-2z_{1}z_{2}+z_{2}^2}
$$
 A: Simply use $$(a-b)^2 = a^2 + b^2 -2ab$$
to reduce this to 
$$\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2}$$
which represents the distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in Euclidean space. 
In terms of a distance vector ${\vec r}$ which points from the first to the second point, this is simply $\sqrt{r^2} = \vert r \vert$. 
A: The function $$f(x_1,x_2,y_1,y_2,z_1,z_2)=\sqrt{x_{1}^2-2x_{1}x_{2}+x_{2}^2+y_{1}^2-2y_{1}y_{2}+y_{2}^2+z_{1}^2-2z_{1}z_{2}+z_{2}^2}$$ is not a polynomial or even a rational function (over the real numbers). Suppose it was a rational function. Then $$g(x)=f(x,0,0,0,0,0)=\sqrt{x^2}=|x|$$ would also have to be a rational function - of a single variable. So let's write it as $g(x)=\frac{P(x)}{Q(x)}$. Observe that $Q$ has no (real) zeros, so it is either always positive or always negative. (We may assume the former, if needed.) The relation $$P(x)=|x|Q(x)$$ holds for all $x\in\mathbb R$. But then $$R(x)=|x|Q(x)-xQ(x)$$ is a nonconstant polynomial with infinitely many zeros, contradicting the fundamental theorem of algebra.
A: This is just:
$$\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2}$$
I don't think there is a nicer way to represent it than this.
A: The simple answer would be that you can't.
As you can see, the expression inside the square root is:
$d^2=(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2$
You might want to know that $d$ is the distance between the point with coordinates $(x_1,y_1,z_1)$ and the one with coordinates $(x_2,y_2,z_2)$
Now, this is a tangent to your question, but if you build a system of spherical coordinates centered at the first point, your expression would be $r$. If you don't get what $r$ is, check this:
http://en.wikipedia.org/wiki/Spherical_coordinate_system
A: Using vector notation, let $\vec r_1=\hat x_1x_1 +\hat y_1y_1+\hat z_1z_1$ and $\vec r_2=\hat x_2x_2+\hat y_2y_2+\hat z_2z_2$.
Define the vector $\vec r$ as the vector from $\vec r_1$ to $\vec r_2$.  Then, the magnitude of this vector is
$$\begin{align}
|\vec r|&=|\vec r_2 -\vec r_1|\\
&=\sqrt{(x_2 -x_1)^2+(y_2 -y_1)^2+(z_2 -z_1)^2}\\
&=\sqrt{x_2^2+2x_1x_2+x_1^2+y_2^2+2y_1y_2+y_1^2+z_2^2+2z_1z_2+z_1^2}
\end{align}$$
So, the term of interest can be written as $r =|\vec r|=|\vec r_2 -\vec r_1$.|
