Why do some people write the gravitational force as proportional to $\mathbf{r}/|\mathbf{r}|^3$? I'm reading Mathematical Aspects of Classical and Celestial Mechanics, Second Edition by Arnold, Kozlov, and Neishtadt. It occurred to me that many people like to use third power when mention the law of universal gravitation. 
For example,

$$F_{ij}=\frac{\gamma m_i m_j}{|r_{ij}|^3}r_{ij}$$ where $r_{ij}=r_i-r_j, \gamma =\text{const}>0$ 

Is this just some conversion/tradition/habit, or does it carry any significance? i.e. the absolute value changed sign and derivative?
 A: It is a matter of convenience:
Let $\vec{r_{ij}}$ be a distance vector with magnitude $r_{ij}$ along the line connecting the masses $m_i$ and $m_j$. Then:
(I): $\vec{r_{ij}}$ squared is a scalar whose value equals its magnitude squared. Proof: $\vec{r_{ij}}^2=\vec{r_{ij}} \cdot \vec{r_{ij}}=r_{ij} r_{ij} \cos{0}=r_{ij}^2$.
(II): $\vec{r_{ij}}$ can be written as $\vec{r_{ij}}=r_{ij} \hat{r_{ij}}$, with $\hat{r_{ij}}$ being a unit vector having the same direction as $\vec{r_{ij}}$. Proof: This goes as an axiom ;-)
If ones writes Newton’s law of gravity as
$$F_{ij}=\frac{γm_im_j}{\vec{r_{ij}}^2}=\frac{γm_im_j}{r_{ij}^2} \tag{1}$$
Then (1) is an incomplete description of the gravitational force. Equation (1) only represents the magnitude of the gravitational force, as can be noticed by the fact that the right side of it is just a scalar.
It is a well-established observation that gravity is a force whose direction is along the line connecting the masses $m_i$ and $m_j$. As a force, gravity must be written as a vector, and therefore the right form would be
$$\vec{F_{ij}}=\frac{γm_im_j}{r_{ij}^2} \hat{r_{ij}} \tag{2}$$
From (II), equation (2) can be written as
$$\vec{F_{ij}}=\frac{γm_im_j}{r_{ij}^2} \frac{\vec{r_{ij}}}{r_{ij}}=\frac{γm_im_j}{r_{ij}^3} \vec{r_{ij}} \tag{3}$$
Typing vectors and unit vectors with arrows and “hats” atop (as in current case) is a little cumbersome. To solve for typing vectors with arrows, one can select other options as boldfacing them for example. To avoid unit vectors one uses the convention $\frac{\vec{r}}{r}$, i.e., the choice of (3) is a convenient form of writing these sort of equations, especially for works containing a huge amount of text. 
It is not only easier (if one uses a WYSIWYG text editor of typing machine) to type equation (3) using these conventions, but visually cleaner as well. Just check:
\begin{equation}
\mathbf{F}_{ij} \boldsymbol{=}\mathrm γ\dfrac{m_i m_j}{\mathbf{r}_{ij}^3 } \mathbf{r}_{ij}
\tag{4}  
\end{equation}
As was pointed out by @AndrewSteane, there are even mathematical advantages by using this notation, hence, it is just more convenient to use the $\frac{\bf r}{r^3}$ notation.
The problem with this, however, is that people (almost) always is introduced to Newton’s law of gravity in its non-vector form (equation 1) and is expecting to find the inverse square law in a first glance, therefore, it may look odd for beginners until one gets used to it some point in future ;-)
A: This is not about gravity but about making maths easier; the same idea comes up whenever you have a force in the radial direction. The reason that ${\bf r}/r^3$ is a little easier to work with than $\hat{\bf r}/r^2$ is the following.
First, one can write it out in rectangular coordinates easily:
$$
\frac{\bf r}{r^3} = \frac{1}{r^3} \left( \begin{array}{c} x\\y\\z \end{array} \right)
$$
Secondly, because $\hat{\bf r}$ has a square root hidden in it which makes some manipulations trickier:
$$
\hat{\bf r} = (x^2 + y^2 + z^2)^{-1/2}(x {\bf i} + y {\bf j} + z {\bf k})
$$
Indeed, when doing things like differentiation, the first step is often to write $\hat{\bf r}$ as ${\bf r}/r$.
Overall, the move to replace $\hat{\bf r}$ by ${\bf r}/r$ reduces the number of different symbols in the mathematical expression and that is the main reason why it is regularly done. Having completed some algebra, one may then choose to present the result in terms of $\hat{\bf r}$ so as to draw attention to the overall scaling of the result, for example in order to make it clear that the gravitational law is an inverse square not an inverse cube law. 
A: 
\begin{equation}
\mathbf{F}_{ij} \boldsymbol{=}\mathrm k\dfrac{m_i m_j}{\Vert \mathbf{r}_{i}\boldsymbol{-} \mathbf{r}_{j}\Vert^2 } \mathbf{n}_{ij}
\tag{1}\label{1}  
\end{equation}
where $\:\mathbf{n}_{ij}\:$ the unit vector along the vector $\:\left(\mathbf{r}_{i}\boldsymbol{-} \mathbf{r}_{j}\right)$. But
\begin{equation}
\mathbf{n}_{ij} \boldsymbol{=}\dfrac{\left(\mathbf{r}_{i}\boldsymbol{-} \mathbf{r}_{j}\right)}{\Vert \mathbf{r}_{i}\boldsymbol{-} \mathbf{r}_{j}\Vert }
\tag{2}\label{2}  
\end{equation}
so the third power
\begin{equation}
\mathbf{F}_{ij} \boldsymbol{=}\mathrm k\dfrac{m_i m_j}{\Vert \mathbf{r}_{i}\boldsymbol{-} \mathbf{r}_{j}\Vert^3 } \left(\mathbf{r}_{i}\boldsymbol{-} \mathbf{r}_{j}\right)
\tag{3}\label{3}  
\end{equation}
