Why does Newton's law of gravitation involve a cube on the bottom?

[Calpalned]

1. The problem statement, all variables and given/known
according to my calculus III textbook, the gravitational force acting on an object at $x = <x,y,z>$ is $F(x)=-\frac{mMG}{|x|^3} x$. What's the point of having a cube on the bottom. Why shouldn't I memorize it as $F(x)=-\frac{mMG}{|x|^2}$

Homework Equations

See above

The Attempt at a Solution

 

[Calpalned]

1. The problem statement, all variables and given/known
according to my calculus III textbook, the gravitational force acting on an object at $x = <x,y,z>$ is $F(x)=-\frac{mMG}{|x|^3} x$. What's the point of having a cube on the bottom. Why shouldn't I memorize it as $F(x)=-\frac{mMG}{|x|^2}$

Homework Equations

See above

The Attempt at a Solution

 

[gneill]

x is a vector. $G\frac{M m}{|\mathbf{x}|^2}$ is the magnitude of the force, while $\frac{\mathbf{x}}{|\mathbf{x}|}$ is a unit vector in the direction of the force.