With stresses acting on a body, they are in what's called rotational equilibrium. What this means is the moments about an origin on a body cancel out such that angular acceleration is zero. This is in fact why our stress matrix
[ 1 ]
$$ [\sigma] = \begin{bmatrix} \sigma_x & \tau_{xy} & \tau_{xz} \\ \tau_{yx} & \sigma_y & \tau_{yz} \\ \tau_{zx} & \tau_{zy} & \sigma_z \end{bmatrix} $$
is symmetric. Because we are in rotational equilibrium, we can say the summation of moments is equal to zero, or
[ 2 ]
$$ \Sigma M = 0 $$
and because of this, we can conclude that $ \tau_{xy} = \tau_{yx} $, $ \tau_{yz} = \tau_{zy} $, and $ \tau_{xz} = \tau_{zx} $. Hence, there are only 6 individual stress states.
Now let's take a look at the cube in a 3D coordinate system that is fixed on 3 sides and has stresses acting upon it below.
Fig. 1 - Image of a cube being acted upon by stresses.
Now Figure 1 may look chaotic at first, but let's break it down. We have a cube (in purple) set in a 3D coordinate system. There are 3 faces that are fixed in space; these are the faces with dotted stresses on them. On the other 3 faces, notice the nasty looking stresses with a stress plus a partial derivative of the stress with respect to the axes the plane is perpendicular to. Essentially what these terms are is accounting for the stress at the fixed end, plus the change in stress as the cube is deformed.
I have highlighted three stresses in green that point in the negative x direction and three stresses in blue that point in the positive x direction. For the forces in the negative x direction, the sum of them can be written as
[ 3 ]
$$ \sigma_{11} dx_2 dx_3 + \sigma_{21} dx_1 dx_3 + \sigma_{31} dx_1 dx_2 $$
Notice we took each stress pointing in the negative x direction and multiplied by two infinitesimal changes. These infinitesimal changes were determined by the axes that needed to be traveled along to reach the stress. For example, $ \sigma_{11} $ is located on the place in the $ x_2 $ and $ x_3 $ frame, hence, we multiply it by $ dx_2 $ and $ dx_3 $ to account for any change it may have in those two directions. We do the same for the positive x axis stresses to get
[ 4 ]
$$ (\sigma_{11} + \frac{\partial \sigma_{11}}{\partial x_1} dx_1) dx_2 dx_3 + (\sigma_{12} + \frac{\partial \sigma_{12}}{\partial x_2} dx_2) dx_1 dx_3 + (\sigma_{13} + \frac{\partial \sigma_{13}}{\partial x_1} dx_3) dx_1 dx_2 $$
Now, if we set these equal to one another, we get some cancellations. We can subtract the $ \sigma_{11} dx_2 dx_3 $, $ \sigma_{11} dx_2 dx_3 $, and $ \sigma_{11} dx_2 dx_3 $ from both sides and then divide by $ dx_1 $, $ dx_2 $, and $ dx_3 $ to get
[ 5 ]
$$ \frac{\partial \sigma_{11}}{\partial x_1} + \frac{\partial \sigma_{21}}{\partial x_2} + \frac{\partial \sigma_{31}}{\partial x_3} = 0 $$
Now Eq. (5) must be true at all times for a stress field to be valid! In fact, this process can be repeated for the y and z directions to get a system of three equations; these are called the Equilibrium Equations for Stresses.
[ 6 ]
$$ \frac{\partial \sigma_{xx}}{\partial x} + \frac{\partial \tau_{yx}}{\partial y} + \frac{\partial \tau_{zx}}{\partial z} + f_x = 0 $$ $$ \frac{\partial \tau_{xy}}{\partial x} + \frac{\partial \sigma_{yy}}{\partial y} + \frac{\partial \tau_{zy}}{\partial z} + f_y = 0 $$ $$ \frac{\partial \tau_{xz}}{\partial x} + \frac{\partial \tau_{yz}}{\partial y} + \frac{\partial \sigma_{zz}}{\partial z} + f_z = 0 $$
I swapped out the shear stresses for $ \tau $'s and replaced 1, 2, and 3 for x, y, and z, respectively because they mean the same thing notation-wise. I also added an $ f_x $, $ f_y $, and $ f_z $ to account for additional body forces such as gravity or magnetic fields. These can be assumed to be zero, but it is always good to check first!
If you are given a stress field, your first instinct should be to verify that it satisfies the Equilibrium Equations for Stresses! If it does not, then the stress field is not physical.