Ah, the cube. EE students are tortured with it since 1826. If you know how to solve it it is easy 
I can now see it's a cube and the simulator told me it's 5 Ohms but I wouldn't know how to get there.
It is not only a cube, but A and B are on opposite ends. So everything is symmetrical. This is the key.
Start at A. Because of the symmetry, the currents in R1, R4 and R12 (those directly connected to A) must be the same. We don't need to know how large they are, just that the three currents are the same. And because R1, R4 and R12 are each 6 Ohm, Ohm's law tells us there is the same voltage across them.
Now for the next trick. Because there is the same voltage over R1, R4, and R12 their other ends (the ends not connected to A) are at the same voltage potential. Because of this one can connect these other ends together. It doesn't change a thing, because of the same voltage potential there will be no current over these added connections. But these connections parallel the three resistors, and one can replace them with a single one. Now comes into play that the professor did use 6 Ohm for each resistor. It makes calculation without a pocket calculator easy. 6 Ohm || 6 Ohm || 6 Ohm = 2 Ohm.
Do the same for the resistors at B, R5, R7 and R8.
Then you see that R2, R3 are also parallel. Replace them with R2 || R3 = 3 Ohm. Same for R9, R10.
The rest should be straight-forward. Sum up the resistors and you should again get two parallel ones. I would now need to make a drawing, but I am to lazy to do it.
Another way to solve this is to think of the cube, cut it along a plane through A-B into two pyramids, calculate the pyramids and parallel the resulting pyramid resistors. When cutting the cube, the resistors at the vertices you cut through need to be split into two 12 Ohm parallel resistors.
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