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Is there any way to simplify the following expression into a non-radical form? $$ \sqrt{x_{1}^2-2x_{1}x_{2}+x_{2}^2+y_{1}^2-2y_{1}y_{2}+y_{2}^2+z_{1}^2-2z_{1}z_{2}+z_{2}^2} $$

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    $\begingroup$ this is the 420,000st question $\endgroup$ – steedsnisps Apr 6 '15 at 17:40
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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

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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 $\sqrt{r^2} = \vert r \vert$.

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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.

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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.

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  • $\begingroup$ This was my answer as well - I don' believe there is any better way to represent it. $\endgroup$ – Alfred Yerger Apr 6 '15 at 17:37
  • $\begingroup$ You can write it down as $|v_1 -v_2|$ where $v_i \in r^3$ which might be a little briefer, but yeah.. :) $\endgroup$ – Guest 86 Apr 6 '15 at 17:37
  • $\begingroup$ +1 because my answer is not very different, so it is unfair for it to be sitting at +3 while yours sits at +1. Cheers :) $\endgroup$ – 299792458 Apr 6 '15 at 17:47
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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 $r =|\vec r|=|\vec r_2 -\vec r_1$.|

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