Counting spanning trees when blowing up vertices I have a cubic graph $G$ with $\tau(G)$ spanning trees. Now I replace each vertex in $G$ by a triangle giving me a new graph $G'$ - this operation is sometimes referred to as blowing up the vertex to a triangle or truncating the vertex. I want to find a formula for the number of spanning trees in $G'$.
For example, the complete graph on 4 vertices $K_4$ has 16 spanning trees, and blowing up each vertex in $K_4$ gives a graph on 12 vertices with 6000 spanning trees. If it is not possible to give a formula for blowing up the vertices in a general cubic graph, a formula for the specific case where the starting graph is $K_4$ is much appreciated.
As an example of how the operation works, I have provided a drawing of the graph obtained from performing the operation once on $K_4$.
(source)
 A: The resulting graph is the linegraph of the subdivision graph of $G$.  This survey paper of Bojan Mohar tells how to obtain the Laplacian spectrum of the linegraph of a semiregular graph and the subdivision graph of a regular graph.
Let's generalize.  Let $G$ be a regular graph of $n$ vertices, degree $d$, and therefore $m=nd/2$ edges.  Let $\mu(G,x)$ denote the characteristic polynomial
of the Laplacian matrix, and let $\kappa(G)$ be the number of spanning trees.
The blowup $B(G)$ of $G$, formed by replacing each vertex by a $d$-clique, is
the linegraph of the subdivision graph of $G$.  Using Theorems 3.8 and 3.9 in the
survey paper of Mohar, we find
$$ \mu(B(G),x) = (-1)^n (x-d)^{m-n} (x-d-2)^{m-n} \mu(G,x(d+2-x)). $$
We know that $\kappa(G) = n^{-1} (-1)^{n-1} \mu'(G,0)$.
Differentiating and using $\mu(G,0)=0$, we find 
$$ \kappa(B(G)) = d^{m-n-1} (d+2)^{m-n+1} \kappa(G). $$
For the $k$-fold blowup, we have
$$ \kappa(B^k(G)) = d^{d_k(m-n)-k} (d+2)^{d_k(m-n)+k} \kappa(G), $$
where $d_k=1+d+\cdots+d^{k-1}$.
For $d=3$, I believe this gives
$$ \kappa(B^k(G)) = (5/3)^k 15^{(3^k-1)n/4} \kappa(G). $$
