Suppose we are given a cube and we add a pair of crossing edges inside each of its faces. It is clear that this drawing has 6 crossings. My question is whether such a graph has crossing number 6? How to prove or disprove it?


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[What is the crossing number of dodecahedron with a copy of $K\_5$ inside each face](https://mathoverflow.net/questions/375927/what-is-the-crossing-number-of-dodecahedron-with-a-copy-of-k-5-inside-each-fac)



What is the crossing number of cube with a pair of crossing edges inside each face
Yes, the crossing number is $6$.


In general if a graph with $V$ vertices and $E$ edges
can be drawn without crossings then $E \leq 3V-6$,
by a familiar application of Euler's formula $V-E+F = 2$:
there are $F = E-V+2$ faces; but each edge separates $2$ faces
while each face has at least $3$ sides; so
$$
2E \geq 3F = 3(E-V+2) = 3E - (3V-6),
$$
whence $E \leq 3V-6$ as claimed.


Now if the graph can be drawn with $c$ crossings then
we can remove $c$ of the edges to obtain a graph with
$V$ vertices, $E-c$ edges, and a crossing-free drawing.
Therefore $c \geq E - 3V + 6$.


Your graph has $V=8$ and $E=24$, so $c \geq 24 - 3 \!\cdot\! 8 + 6 = 6$, **QED**.


