Seven cubes, whose volumes are $1$, $8$, $27$, $64$, $125$, $216$, and $343$ cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area of the tower (including the bottom) in square units?
The side length of these volumes are $1$, $2$, $\cdots$, $7$. Hence, the surface are of them are $6\times 1^2$, $6\times 2^2$, $\cdots$, $6\times 7^2$. This meas that the total surface area without being covered is $$\sum_{n=1}^{7}\left(6\times n^2\right)= 6\times \frac{7\times (7+1)\times (2^7+1)}{6}=840$$
When the are stacked in the described way, the bottom face of a smaller volume will be covered by the volume beneath it. This results a loss of two times of the surface area of the smaller volume's bottom face. Therefore, the total loss equals: $$\sum_{n=1}^6\left(2\times n^2\right)=2\times \frac{6\times (6+1)\times(2\times 6 + 1)}{6}=182$$
It follows that the final answer is $$840-182=\boxed{658}$$