I1.1 Three cubes with volumes $1,8,27$ are glued together at their faces. If $a$ is the smallest possible surface area of the resulting polyhedron, find the value of $a$.
The lengths of the 3 cubes are 1,2 and 3 with surface areas 6 , 24 and 54 respectively.
As shown in the figure, if the three cubes are glued together, the faces stuck together are $2 \times 2,2 \times 2,1 \times 1,1 \times 1,1 \times 1$ and $1 \times 1$. The smallest possible surface area is $6+24+54-4-4-4$. $a=72$