A cube with an edge of $12 \mathrm{~cm}$ was divided into smaller, identical cubes such that the sum of the surface areas of all the smaller cubes was eight times greater than the surface area of the original cube.

Determine how many small cubes there were and what the length of their edges was.

(M. Volfová)
If the edges of the smaller cubes are $n$ times smaller than the edge of the original cube, then the original cube was divided into $n \cdot n \cdot n=n^{3}$ smaller cubes. The surface area of the cube is $6 \cdot 12^{2}\left(\mathrm{~cm}^{2}\right)$. The sum of the surface areas of all the smaller cubes is

$$
n^{3} \cdot 6 \cdot\left(\frac{12}{n}\right)^{2}=n \cdot 6 \cdot 12^{2}\left(\mathrm{~cm}^{2}\right)
$$

For this sum to be eight times larger than the surface area of the original cube, $n$ must be 8. The cube was divided into $8^{3}=512$ smaller cubes, and the edge of each of them measured $12: 8=1.5(\mathrm{~cm})$.

Evaluation. 2 points for the relationship between the ratio of the lengths of the edges of the cube and the smaller cubes $(1: n)$ and the number of all smaller cubes $\left(n^{3}\right) ; 2$ points for expressing and comparing the surface areas; 2 points for solving the problem.

Note. The same result can also be achieved by progressively trying to divide the cube, which leads to $\mathrm{k}$ calculations corresponding to substituting $n=2,3, \ldots$ into the previous expressions. $\mathrm{V}$ In such a case, adjust the evaluation based on the quality of the commentary.