Cross terms appear when the coordinate axes do not pass through the center of mass. That is if you start with a diagonal inertia matrix at the center of mass, when applying the parallel axis theorem cross terms will appear.
In vector form the parallel axis theorem is
$$ {\bf I} = {\bf I}_{cm} - m [{\bf r}\times] [{\bf r}\times] $$
where $[{\bf r}\times] = \begin{pmatrix}x\\y\\z\end{pmatrix} \times = \begin{bmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{bmatrix}$ is the cross product matrix operator.
So if we start with a diagonal inertia at the center of mass, when moved to a different point $(x,y,z)$ the inertia matrix is
$$ {\bf I} = \begin{bmatrix}
I_x+m(y^2+z^2) & - m x y & -m x z \\
- m x y & I_y + m (x^2+z^2) & - m y z \\
- m x z & - m y z & I_z + m (x^2+y^2)
\end{bmatrix} $$
So if any two of $x$ $y$ or $z$ are zero the result is still a diagonal matrix. This happens when one of the coordinate axis passes through the center of mass. In your case if the $x$ axis goes from the corner towards the center of mass (across diagonal) then $y=0$ and $z=0$ and the criteria is met.
So the question boils down to under which conditions the inertial matrix is diagonal at the center of mass. The answer to this has to do with symmetries. For example, when a particle at a positive $z$ coordinate is matched with an particle at a negative $z$ coordinate then when the cross terms get added up the result is zero. This can be seen from the structure of the inertia matrix (as shown above) if you think of this as a parallel axis theorem for a lot of small little point masses added together.
It is simply not true, as stated.
A cube is as perfectly balanced around its center of mass as a sphere is. You have shown it mathematically, and you are perfectly correct.
However, the principal axes may be chosen perpendicular to the faces of the cube. Of course, when the point is the center of mass, this is no better than any other choice, except for considerations of symmetry and mathematical convenience.
Best Answer
1) I don't understand your questions really well. Consider rephrasing it to make it a bit clearer. What I can tell you is the following:
The mathematical definition of the principal axes of inertia is that they are the eigenvectors of the inertia tensor.
The physical interpretation of the principal axes of inertia is that they represent the directions along which the angular momentum $\mathbf{L}$ is parallel to the angular velocity $\mathbf{\omega}$, so $\mathbf{L} = I\mathbf{\omega} $ with $I$ = constant as opposed to $\mathbf{L} = \underline{\underline{I}} \cdot \mathbf{\omega} $.
When calculating the inertia tensor $\underline{\underline{I}} $, you use a certain basis (so that you find all the distances between the parts of the system). If that basis were to be the set of principal axes, then the matrix would come out to be diagonal. In other words, if you want to diagonalise the inertia tensor via a similarity transformation $\underline{\underline{S}}^{-1}\underline{\underline{I}}\underline{\underline{S}}$, you would have to use the matrix made of the principal axes of inertia as $\underline{\underline{S}}$.
2) There are some tricks to determine which axes are principal axes. They can be shown mathematically but usually a qualitative understanding is enough (i.e. it makes sense - why would they not be?).
If an object has a rotational symmetry (i.e. you can rotate about an axis without changing anything about it - shape, mass distribution etc.) then the axis of symmetry is a principal axis.
If an object has a reflection/mirror symmetry (i.e. an equilateral triangle is symmetric about a line/plane through a vertex and the mid-point of the opposite edge), then the direction normal to the plane of symmetry is a principal axis.
3) In the case of a diagonal inertia tensor (i.e. using the principal axes as the basis), $I_{xx} = I_{yy}$ means that two eigenvalues are the same and therefore they are degenerate. If you try and calculate the eigenvector associated with this eigenvalue you will find a plane (an equation in terms of x, y and z). Basically any vector on that plane will be a principal axis, but it is customary to choose two orthonormal vectors, just to make life easier.