General idea behind simplifying cube resistors

If you are trying to find the resistance between two points, you suppose that a pd is placed between these points. You then look for symmetries that show certain points to be at the same potential. When you find such points, you imagine them to be linked by wires of negligible resistance, knowing that there will be no current in these wires, so their presence won't affect the functioning of the circuit. But linking the points at the same potential will usually reveal a much simpler equivalent circuit.

For example, suppose you are trying to find the resistance between two points at opposite ends of a body diagonal, AG, of a cube with equal resistances along each edge. [Two opposite faces of the cube are ABCD and EFGH with the other four edges of the cube being AE, BF, CG and DH.]

A is linked by single resistors to B, E and D. The ways that current can get from these points to G, involve identical resistor combinations [e.g. current from B can go to G via C or F, and you should check the routes to G from E and D.] Thus B, E and D must be at equivalent points in identical potential dividers, and are all at the same potential. If we link B, E and D by wires of zero resistance, there will be no current in these wires. The same goes for G's 'neighbours', points F, C and H.

So imagine these wires to be in place. You'll then see that the 'resistor cube' with 'terminals' at A and G is equivalent to 3 resistors in parallel, in series with 6 in parallel and another 3 in parallel.

You can find the resistance between A and E, A and F and possibly other pairs of points by using variations of this method. Redrawing the cube as a two-dimensional circuit may (or may not) help.

The sure-fire guaranteed method, which works even if the 3D network cannot be drawn as a 2D network so that the series and parallel formulas can be used, and even if there is no symmetry to give helpful equipotentials, as in the case with a cube with 12 different resistors along its edges, is:

You apply a voltage $V_0$ between nodes A and B.

Define a current $I_j$, with $j$ from 1 to 12, through every resistor. (Be very careful about directions and signs. Drawing a picture with arrows helps)

The voltage through each resistor $j$ is then $I_j R_j$.

Apply Kirchoff's first law to every node other than A and B: current in = current out. This gives 6 equations for the 12 unknown currents.

Trace all possible independent paths from A to B and apply Kirchoff's 2nd Law:
the total voltage must be $V_0$. This gives another 6 equations.

This gives 12 equations in 12 unknowns, which can be solved. So you know all the currents (in terms of $V_0$) and can find the current flowing into A, or equivalently out of B. The ratio of $V_0 / I_{total}$ gives the resistance.

You can define loop currents which give a short cut by automatically satisfying Kirchoff's 1st law, if you want, at the price of more complicated expressions in the second law.

This works for networks other than the cube. But it's complicated and you can see why people use shorter methods if the situation is simple enough for them to do so.