[Geometry]
Problem: Let $T=8$. A cube has volume $T-2$. The cube's surface area equals one-eighth the surface area of a $2 \times 2 \times n$ rectangular prism. Compute $n$.

Solution: The cube's side length is $\sqrt[3]{T}$, so its surface area is $6 \sqrt[3]{T^{2}}$. The rectangular prism has surface area $2(2 \cdot 2+2 \cdot n+2 \cdot n)=8+8 n$, thus $6 \sqrt[3]{T^{2}}=1+n$. With $T=8, n=6 \sqrt[3]{64}-1=\mathbf{2 3}$.

Final answer: 23