Consider a cube of 1 Ohm resistors... What is the resistance between 2 points in a cube of 1-ohm resistors? I am asking about each of the cases:
1) The 2 points are connected by 1 edge
2) The points are 2 edges from each other
3) The points are antipodal (if such a word can be used in the case of a cube).
 A: A MATLAB solution:
% Study a network of twelve 1-ohm resistor connected to form a cube.
% We want to compute the resistances between pairs of nodes.
%
% The vertices of the cube are numbered in binary from 000 to 111.
% We want to compute the resistance between nodes 001 and 000 (adjacent), 
% 011 and 000 (across a face), and finally 111 and 000 (opposite).

% The voltage of vertex 000 is taken to be 0.
% Writing the node-voltage equations for nodes 1-7 we get:
A = [ 3  0 -1  0 -1  0  0;
      0  3 -1  0  0 -1  0;
     -1 -1  3  0  0  0 -1;
      0  0  0  3 -1 -1  0;
     -1  0  0 -1  3  0 -1;
      0 -1  0 -1  0  3 -1;
      0  0 -1  0 -1 -1  3];
% The three queries result in the following right-hand sides:
B = [1 0 0; 
     0 0 0; 
     0 1 0;
     0 0 0;
     0 0 0;
     0 0 0;
     0 0 1];
% We solve symbolically to get fractions.
S = sym(A) \ sym(B);
disp(['The resistance between nodes 001 and 000 is ', char(S(1,1))])
disp(['The resistance between nodes 011 and 000 is ', char(S(3,2))])
disp(['The resistance between nodes 111 and 000 is ', char(S(7,3))])

which produces:
The resistance between nodes 001 and 000 is 7/12
The resistance between nodes 011 and 000 is 3/4
The resistance between nodes 111 and 000 is 5/6

A: We can solve this problem without applying Kirchoff's laws at all, just by exploiting symmetry until we're left with series-parallel graphs.
Label the cube as follows:

If we want the effective resistance between $a$ and $b$, then we know $V_c = V_e$ and $V_d = V_f$, because everything is symmetric through the plane $abgh$, so we can contract these to two points $ce$ and $df$:

Now we can use the series-parallel rules to write the effective resistance between $a$ and $b$ as
$$\frac{1}{\frac{1}{1} + \frac{1}{\frac12 + \frac1{\frac1{1/2} +\frac1{1/2+1+1/2}} + \frac12}} = \frac7{12}.$$
If we want the effective resistance between $a$ and $g$, we have to be slightly more clever. If $V_a = 0$ and $V_g = 1$, then $V_c = V_d = V_e = V_f = \frac12$ because the cube is symmetric through plane $cdef$, but these points are equidistant from $a$ and $g$. So we can contract all four of these to one point:

Now we can use the series-parallel rules to write the effective resistance between $a$ and $g$ as $$\frac{1}{\frac{1}{1/2} + \frac{1}{1 + 1/2}} + \frac{1}{\frac{1}{1/2} + \frac{1}{1 + 1/2}} = \frac34.$$
Finally, to get the effective resistance between $a$ and $h$, we can use rotational symmetry about the diagonal $ah$ to collapse $bce$ to one point and $dfg$ to another:

Here, the effective resistance is just $\frac13 + \frac16 + \frac13 = \frac56$.
There's also a shortcut to solving the adjacent-vertices case. If $R_{vw}$ is the effective resistance between vertices $v$ and $w$, then $$\sum_{vw \in E} R_{vw} = n-1,$$ where $vw \in E$ ranges over all edges of the graph, and $n$ is the number of vertices. In this case, all $12$ edges have equal effective resistance by symmetry, and these add up to $n-1 = 7$, so the resistances must be $\frac{7}{12}$ each.
