"Resistor cube" - what is the basic approach?

This might be a more intuitive approach. It takes advantage of the symmetry (if you can find it) of the circuit.

Figure 1. Original drawing with one of the 'G' nodes replaced by 'E'.

^{simulate this circuit – Schematic created using CircuitLab}

Figure 2. Original circuit drawn in more conventional format. Now the symmetry becomes apparent.

^{simulate this circuit}

Figure 3a. Applying a star to delta transform on the star connected resistors at node F. 2b. Since both sides of R12 are at the same potential (due to the circuit's symmetry) no current will flow it it so it can be removed without affecting the overall resistance.

^{simulate this circuit}

Figure 4. Taking advantage of the circuit's symmetry we can split it in half vertically into to parallel circuits.

It should now be a fairly simple exercise to calculate R_AB.

Slice your cube with a vertical diagonal plane to 2 exactly equal pieces. Remove exactly a half of the conductive material and current between A and B. Then you have left this:

This is a simple circuit. Its resistance between A and B is 14R/5. One gets it by using the series and parallel resistance formulas.

The actual resistance of the original cube is a half of this.

ADD: Why this works? Because the slicing plane happens to be a perfect mirror symmetry plane for the current flow field. When a source is connected between A and B, no current goes through the plane

• in joints A and B,
• in the ends of the sliced resistor
• inside the sliced resistor

Thus the load seen by the source can be seen as 2 identical loads in parallel; there's no intermediate current paths in use between the halves.

Fortunately I happened to see the symmetry. The circuit has too many nodes and current loops for assuming some non-zero voltage or current source connected between A and B and using KCL or KVL equations written with no "keep it under law and order" mechanism. Circuit analysis programs can do it and perhaps some math program could parse and solve the irregularly written equations. For a practical electrician like me the task is quite hopeless.

I will focus on your question

Where did I err, and without knowing a voltage, how do you approach this?

rather than giving a detailed solution, which the other answers already do nicely.

Your circuit does not include any current or voltage source - the solution you get is correct: clearly, there will be no currents flowing through any of the resistors in the absence of a source.

When calculating the equivalent resistance of the network, the implicit question being asked is: "If a voltage source is connected between nodes A and B, what will be the ratio \$V/I\$ between its voltage \$V\$ and the current \$I\$ flowing into the resistor network?". After adding the voltage source, an additional loop current \$I\$ (through AGHB) needs to be added and the corresponding voltage equation around the loop will involve \$V\$. When you solve the system of 5 equations for \$I_1, I_2, I_3, I_4, I\$, you will get a non-zero solution this time (because the right-hand side of your equations contains \$V\$ instead of being a zero vector). Then, taking the ratio \$V/I\$, you get the equivalent resistance.

I will take a clue from your comment about graph Laplacians (a directed graph replacing an orthogonal field, in effect.)

If you design the directed graph for maximum simplicity, then you will lay out the graph matrix such that the two nodes of interest are the last two. In particular, I'll place your \$A\$ as the 2nd to last node and your \$B\$ as the last node:

^{simulate this circuit – Schematic created using CircuitLab}

This generates an incidence matrix for the above directed graph:

A = Matrix([
  [ 1,  0,  0,  0,  0,  0, -1,  0],    # R1
  [ 0,  1,  0,  0,  0,  0, -1,  0],    # R2
  [-1,  0,  1,  0,  0,  0,  0,  0],    # R3
  [ 0, -1,  1,  0,  0,  0,  0,  0],    # R4
  [ 0, -1,  0,  0,  1,  0,  0,  0],    # R5
  [ 0,  0, -1,  0,  0,  1,  0,  0],    # R6
  [ 0,  0,  0,  0,  1, -1,  0,  0],    # R7
  [ 0,  0,  0,  1,  0, -1,  0,  0],    # R8
  [-1,  0,  0,  1,  0,  0,  0,  0],    # R9
  [ 0,  0,  0, -1,  0,  0,  0,  1],    # R10
  [ 0,  0,  0,  0, -1,  0,  0,  1]])   # R11
  ----------------------------------
    1   2   3   4   5   6   7   8  

You probably already know that the Laplacian (unweighed by Ohm's Law) is just \$L=A^T\:A\$ (the divergence of the gradient of the graph.) But to apply Ohm's Law, we need a conductance matrix, \$C\$, to find \$W=A^T\:C\:A\$. But \$\frac1r\,I=C\$, where \$r\$ is the arbitrary value of each equal-valued resistor. So \$W=A^T\:\frac1r\:I\:A=\frac1r\,L\$. Sticking with the simpler \$L=A^T\:A\$ let's simply remove \$r\$ for now, placing it outside of our initial consideration. That keeps things a bit simpler as we go forward. We can re-apply it, later.

The Laplacian is positive semi-definite. But I can treat node 8 as ground -- meaning that I know the potential (voltage value) there is zero, by definition. Knowing that, we can strike out the row and column associated with it. (Since I made it the last row and column, this makes things easy to do.) The resulting reduced \$L\$ is now symmetric-positive-definite!

Call the node fluxes \$\vec{f}\$ and the node potentials \$\vec{x}\$. Then \$L\:\vec{x}=\vec{f}\$. But this equation can be broken down into this:

^{simulate this circuit}

We are only interested in \$\frac{v}{i}\$, with both \$v\$ and \$i\$ being scalars above. We don't need any of the unknowns present in \$\vec{x}\$. And we already know \$\vec{f}\$ is mostly \$\vec{0}\$, due to KCL. So the above can be greatly simplified:

$$\begin{align*} P\,\vec{x}+Q^T\,v&=\vec{0}&\therefore \vec{x}&=-P^{-1}\,Q^T\,v \\\\ Q\,\vec{x}+R\,v&=i \\\\ Q\left(-P^{-1}\,Q^T\,v\right)+R\,v&=i&&\text{by substitution} \\\\ \left(R-Q\,P^{-1}\,Q^T\right)v&=i&\therefore \frac{v}{i}&=\frac1{R-Q\,P^{-1}\,Q^T} \end{align*}$$

Let's perform all this in Sagemath/SymPy/Python. We already have the incidence matrix \$A\$ from above. So I won't repeat that, but just proceed from there:

L = A.T * A
P = L[0:6,0:6]
Q = L[6:7,0:6]
R = L[6:7,6:7]
1/(R-Q*P.inv()*Q.T)[0,0]
7/5

Now we remember that we used the unweighted Laplacian. So to recover, we just multiply the above by \$r\$. So the answer is \$\frac75 r\$.

(If you need to follow why we multiply, rather than divide, note that the \$\vec{0}\$ meant that solving for \$\vec{x}=-P^{-1}\,Q^T\,v\$ eliminated the \$\frac1r\$ factor entirely, but that solving for \$\left(R-Q\,P^{-1}\,Q^T\right)v=i\$ left off the fact that if we'd used the weighted Laplacian, \$W\$, then the equation should have been \$\frac1r\left(R-Q\,P^{-1}\,Q^T\right)v=i\$ and that therefore \$\frac{v}{i}=\frac{r}{R-Q\,P^{-1}\,Q^T}\$.)

A high level perspective isn't too complicated. When looking at any matrix as a mapping between two spaces, you can imagine that the rows represent the basis vectors for the input (source) space -- called the rowspace -- and that the columns represent the basis vectors for the output (destination) space -- called the columnspace.

When you see \$A^T\,C\,A\,\vec{x}=\vec{f}\$ then you should read the left hand side backwards, proceding right to left:

1. \$\vec{x}\$ is a vector in the node-potential space, which will be the source/input space. These node values are absolute. Call the length of \$\vec{x}\$ as \$n\$, so the dimenionality of this input space is \$n\$. (You know from physics that everything is relative, so to speak, so this is already a problem since absolute values need to be converted somehow into relative values.)
2. \$\vec{e}=A\,\vec{x}\$ maps these vectors in the source/input node-potential space into vectors in the potential-differences-across-resistors/edges space. The resulting vector will have dimension \$m\$. (\$A\$ is an \$m\times n\$ matrix.) So the dimensionality of the result is \$m\$. Now, this makes a lot of sense in terms of relativity -- things are now all relative to each other as potential differences across edges/resistors.
3. \$C\$ is an \$m\times m\$ diagonal conductance matrix, with the conductance of each edge down the diagonal and zeros everywhere else. So \$\vec{w}=C\,A\,\vec{x}\$ imposes Ohm's Law so that we now have the fluxes (currents) along the resistors/edges given the potential differences that were just calculated in step 2. This should make sense, as well. The result is still a vector of fluxes, one for each edge. A vector of length \$m\$ residing in a weighted space that is similar, but now weighted, to the earlier potential-differences-across-resistors/edges space so that we have converted the potential differences into fluxes, now, using Ohm's Law.
4. \$\vec{f}=A^T\,C\,A\,\vec{x}\$ now maps these edge fluxes back into a node space, but the novel idea here is that the edge fluxes are now being summed back into the nodes to get the net flux into or out of each node. (\$A^T\$ is an \$n\times m\$ matrix.) Note that \$W=A^T\,C\,A\$, the weighted Laplacian, is square and maps from a node space of dimension \$n\$ to another node space also of dimension \$n\$. (\$L=A^T\,A\$ is also called the divergence of the gradient or the unweighted Laplacian and \$W=A^T\,C\,A\$ is the weighted divergence of the gradient, after applying Ohm's Law, or the weighted Laplacian.) But where at first \$\vec{x}\$ represented absolute potential values at each node, \$\vec{f}=A^T\,C\,A\,\vec{x}\$ has \$\vec{f}\$ representing the fluxes squirting out of or draining into each node. KCL tells us that this should mostly be zero, without some external source of some kind, of course. So we expect most of \$\vec{f}\$ to be a lot of zeros due to KCL. (But we have to allow for the fact that there may be voltage sources or current sources to alter this for specific nodes.) Further, due to the Rank-Nullity theorem it must be the case that \$\sum f_i=0\$, which doesn't require all of the values to be zero but does require their total sum to be zero.

That's a kind of high level view that looks about like this:

(Take note that mathematicians prefer \$\Delta\$ instead of \$\nabla^2\$ for some good reasons. The literature on this can be confusing, depending on context. The reason for this is that some consider \$\nabla\cdot\nabla\$ to be represented conveniently by \$\nabla^2\$, since it looks right. But mathematicians complain because \$\nabla\cdot\nabla\$ gives scalar values and not a Hessian, which is what \$\nabla^2\$ means to mathematicians. Just be aware of varying terminology and keep track of the context, when reading.)

The above approach is general and doesn't depend upon occasional symmetries or ad-hoc \$\Delta Y\$/\$\Delta *\$ conversions. It's a powerful tool; not only for electronics, but for a wide array of problems in physics, probability theory, wave equations, and quantum mechanics, for example. (Not to mention abstract and applied mathematics.)

I do have a little more information at this answer and at this answer.

The matrix resulting from factoring out R is invertible, so I swiftly and unhelpfully obtain I=0.

The problem here comes from a very obvious fact from using KVL (mesh.) Imagine just three resistors in a triangle as a mesh. Assume for a moment that the voltages at each of the nodes are not the same (if they all were the same then obviously all three edge currents must be zero.) Let's say that the current going from node a to node b causes a one-volt drop. So node b is 1 V less than node a. But all of this current entering node b must now go towards node c, due to KCL requirements. Let's say this also causes a one-volt drop, so it must be that node c is 1 V less than node b. But all of this current entering node c from node b must now be going back to node a. Let's say that causes a one-volt drop, too. Then node a must be 1 V less than node c.

But this is impossible because that means node a must be 3 V less than itself. And that's just not possible.

So the only solution possible is that the current is zero. Any other choice with non-zero resistor values would be a contradiction.

That's why you get the solution you get.

What's the answer? Well, you have to embrace the idea that KCL only applies to one of the three nodes in this triangular circuit. The other two nodes must not obey KCL, that is, at least, without adding something not yet shown in the schematic. There must be some current entering one of those nodes (from elsewhere) and the same current must be exiting the other one of those two nodes (back into elsewhere.)

All this means is that you need to inject a current into one of the nodes and remove that injected current from some other node. Then you can get a useful result. Or, you can assign a zero voltage to one of the nodes, arbitrarily, and assign a different non-zero voltage to one of the other nodes. Then you will find that there will be a current flux into one of those nodes from the voltage source that is equally removed from the other node by the same voltage source. Either approach works to solve your difficulty. But you must somehow avoid the contradiction that would otherwise force all the currents to zero.

Since this is multiple choice it must be easy to solve.

In fact it can be done almost without calculation, by only applying basic principles.

• As others have noted there is symmetry: The nodes G&D as well as D&H will always have same voltage. This is clear from how the current flows from A to B.
• Consequently, we can connect G+D and D+H without altering the total behavior (provided we are only interesting in current from A to B).
• This simplifies the circuit drastically, essentially revealing that the 3D arrangement is just there to add confusion.
• We notice that all resistors with exception of the one between F and E are now parallel circuited. Since parallel circuit of same resistors yields half the value, we arrive at the equivalent circuit
• This is easy solve. The only slightly difficult part is the parallel circuit of R/2 and 2R (==R/2+R/2+R) for calculating the resistance between C/G and D/H. But even this we do not need to work out. Lets start with an estimate and check the provided multiple-choice options.
• As we have a small resistance (R/2) and a rather large one (2R) the total resistance of the parallel circuit is a only little smaller than R/2 (It would be half if both resistances were equal, but the more unlike they are the closer the overall resistance to to the lower one).
• So we have for total resistance between A and B we have R/2 + R/2 + "something a little smaller than R/2".
• In total the result must be between R and 1.5R (and being more close to 1.5R than to R).
• Looking at the provided options, we see that only option C is greater than R, so this must be the solution. And indeed 7/5R is a just little smaller than 1.5R (==7.5/5R) which adds confidence that this is right result.

Very good answers! I will give you my two cents also. Whenever you get an unintuitive network of resistances try to "flatten" it out. Then use series parallel and delta/wye transformations to come to your answer. KVL, KCL should be used as a brute force method and it is very tedious as you can see.