= Solution
The <adjacent-corner resistance of a resistor cube> follows from <Kirchhoff's current law> and cube symmetries. Hold adjacent corners $A=(0,0,0)$ and $B=(1,0,0)$ at potentials $V$ and zero. The two corners $(0,1,0),(0,0,1)$ have a common potential $x$; their neighbours $(1,1,0),(1,0,1)$ have $V-x$. Write the remaining two potentials as $z,V-z$.
At a corner of the first pair and at $(0,1,1)$, <Kirchhoff's current law> gives
$$
3x=V+(V-x)+z,\qquad3z=2x+(V-z).
$$
Thus $4x-z=2V$ and $4z-2x=V$, yielding $x=9V/14$, $z=4V/7$. The supplied current at $A$ is
$$
I=\frac VR+2\frac{V-x}{R}=\frac{12V}{7R},\qquad\boxed{R_{\mathrm{eq}}=\frac VI=\frac{7R}{12}.}
$$
The direct edge between the terminals is one of three current paths leaving $A$; treating the other eleven resistors as a single obvious series chain would miss the branching.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2016-resit-cube-resistor-potentials.png]
{title=Cube network potentials for unit applied voltage between adjacent corners; each edge has the same resistance}
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