Resistance across the body diagonal of a cube of 1-ohm resistors

Step-by-step derivation

1. Inject a current $I$ into corner $A$ and extract it from the opposite corner $G$. Exploit the symmetry: the three edges leaving $A$ are indistinguishable from one another, so each carries $I/3$.
2. Those three currents arrive at three "near" corners. From each near corner, two edges lead onward to the three "far" corners. By the same symmetry each of those six middle edges carries half of $I/3$, that is $I/6$.
3. Each of the three far corners is fed by two middle edges carrying $I/6$ each, so each far corner collects $I/3$, and the three final edges into $G$ each carry $I/3$. Total arriving at $G$ is $3 \times I/3 = I$. ✓ Current is conserved, confirming the distribution.
4. Now walk one path from $A$ to $G$ and sum the voltage drops, using $V = IR$ with $R = 1\ \Omega$ on every edge: $$V_{AG} = \underbrace{\frac{I}{3}(1)}_{\text{first edge}} + \underbrace{\frac{I}{6}(1)}_{\text{middle edge}} + \underbrace{\frac{I}{3}(1)}_{\text{last edge}}.$$
5. Add the three terms over a common denominator of $6$: $$V_{AG} = I\left(\frac{2}{6} + \frac{1}{6} + \frac{2}{6}\right) = \frac{5I}{6}.$$
6. Apply Ohm's law to the whole network: $$R_{eq} = \frac{V_{AG}}{I} = \frac{5}{6} = \mathbf{0.833\ \Omega}.$$
7. The other two cases, for completeness. Across a face diagonal the answer is $3R/4$; across a single edge it is $7R/12$. All three follow from the same symmetry argument, and all three have appeared in exams.