How can the potential on a cube's surface replicate the potential of a point particle? If i have a cube (either hollow, or an insulating solid) and i want its surface have a potential such that it looks like a point particle outside of the box does that mean the exact potential on the cube itself is the same as the potential of a point particle?
 A: Suppose you have a charged cube as shown below,

Say it holds a charge of $Q$ Coulomb.
Now, for this cube to behave like a point charge, you need to go very $far$ away from it such that at a particular position, your cube appears like a point charge.
In this case, the cube $behaves$ just like a point charge with a charge of $Q$ Coulomb.

It is clear that as you keep going far away from the charged cube, it keeps getting smaller. At one point, it appears like a point charge and it also behaves like a point charge.
So, one can use the same equation for potential at any point due to a point charge in the case where your charged cube appears like a point charge. Any charged geometrical shape would act as a point charge when viewed from a point very $far$ away from it.
A: Yes, the uniqueness theorem guarantees it.  If you find a solution in which the potential on the surface of the cube looks just like the potential of a point particle, then the potential outside the cube must be identical to that of a point particle since there is only one unique solution satisfying this this boundary condition.  (Assuming no external charges, of course.)
A: 
if the cube is an insulator or a dielectric, how do I find the surface charge density and the volume charge density?

Since you are working with an insulator, you can have $uniform$ surface and volume charge densities unlike in the case of a conductor( other than a conducting sphere ). The reason is, excess charge cannot redistribute in an insulator as they are not free to do so. There is my advantage! Let me first talk about uniform $surface$ charge density of a cube.
Look at the figure below, 

I have displayed all the sides of a charged cube. Here, I have given it a charge of say $Q$ Coulomb. Since it is an insulator, I have succeeded in establishing uniform charge distribution. The total charge available on the $surface$ of the cube is $Q$ Coulomb. The uniform $surface$ $charge$ $density$ in this case is the charge available per unit area given by $\sigma$
$\sigma = \frac{Q}{A}$ where $A$ is the area available. 
For the cube,
$\sigma = \frac{Q}{6\times a^{2}}$ where $a$ is the side of the cube. 
As a result of uniform charge distribution, $\sigma$ will remain constant. That's because, for any area chosen on the surface of the cube, the value of $Q$ will be such that $\sigma$ remains constant. 
Let $A = A1 + A2$  and the available charge in $A1$ is $Q1$ and that in $A2$ is $Q2$
For uniform surface charge distribution, $\frac{Q1}{A1} = \frac{Q2}{A2} = \sigma$
This result always holds for uniform charge distribution.
Now, let us talk about the uniform volume charge density of the cube.
Look at the diagram below,

Once again, I have given a charge of $Q$ Coulomb to the cube in such a way that the charge is uniformly spread in the volume of the insulating cube.
Now, the $volume$ $charge$ $density$ of the cube is given by $\rho$
$\rho = \frac{Q}{V}$ where $V$ is the volume available.
For the cube, 
$\rho = \frac{Q}{a^{3}}$ where $a$ is the side of the cube.
As a result of uniform charge distribution, $\rho$ remains constant. That's because for any volume chosen in the cube, the value of $Q$ will be such that $\rho$ remains constant.
Let $V = V1 + V2$  and the available charge in $V1$ is $Q1$ and that in $V2$ is $Q2$
For uniform volume charge distribution, $\frac{Q1}{V1} = \frac{Q2}{V2} = \rho$
This result always holds for uniform charge distribution.
I have explained using uniform charge distribution which is the simplest case in order to make things easy to understand.
But, if you are willing to make the charged cube behave like a point charge, you do not have to worry about the charge densities. All you need to do is to go very far away from it. At such a great distance from the cube, things will appear as though you are working with a point charge. Dimensions of the cube no longer matter. All that matters is the charge $Q$.
If I have not answered your question, it means that I have not fully understood your question and I would request you to add a few more details in your question.
Thank you.
