The minimum surface area of $k$ unequal boxes tiling a cube: sharp thresholds, a fault-free law, and a reduction to two dimensions

Let $T(n,k)$ be the minimum total surface area of $k$ axis-aligned boxes with integer sides and pairwise distinct dimension multisets whose union is the cube $[0,n{]}^3$. We determine the column $k=6$ completely: $T(n,6)=8n^2+2n+12$ for $5\le n\le 9$, and $T(n,6)=8n^2+2n+6$ for all $n\ge 10$, together with the exceptional values $T(3,6)=100$ and $T(4,6)=148$. The threshold $n=10$ equals $1+2+3+4$, the least possible sum of four distinct stick lengths, and the general law holds: for every $k\ge 4$ and every $n\ge (k-2)(k-1)\mathrm{/}2$, $T(n,k)=8n^2+2n+2(k-3)$, with thresholds at the triangular numbers. Three structural results support and extend these values. First, a fault-free law: the minimum internal interface of a partition of the cube into six boxes with no fault plane is exactly $2n^2+n$ for all $n\ge 3$ (OEIS A014105), proved by an exact accounting of spanning pieces, floating pieces and cube corners. Second, a reduction theorem: within an explicit range, the three-dimensional problem collapses to a two-dimensional one, $I(n,k)=n^2+W^{\ast }(n,k-1)$, where $W^{\ast }(n,m)$ is the minimum internal wall of a tiling of the $n\times n$ square by $m$ rectangles of pairwise distinct dimensions; the key ingredient is an unconditional slab lemma. Third, a doubling law in the middle regime of the 2D problem: $W^{\ast }(n,4)=n+4$ for $n=4,5$ and $W^{\ast }(n,5)=n+6$ for $4\le n\le 9$, proved by finite case trees; via the reduction theorem this gives computer-free proofs of the middle regimes of the columns $k=5$ and $k=6$. The lower bound for the main family does not use the distinctness of the pieces.