Caen has a cube with a volume of \(n \text{ cm}^3\). They cut this cube into \(n\) smaller cubes, each with a side length of \(1\) cm. The total surface area of the \(n\) smaller cubes is ten times the surface area of Caen’s original cube. Determine the side length of Caen’s original cube.
Let the side length of Caen’s original cube be \(x\) cm, where \(x>0\). It follows that \(n=x^3\).
Each of the six sides of Caen’s original cube has area \(x^2\) cm\(^2\), so the total surface area of the original cube is \(6x^2\text{ cm}^2\).
Consider one of the smaller cubes. The area of one the six faces is \(1\text{ cm}^2\). So, the surface area of one of these smaller cubes is \(6\text{ cm}^2\). Thus, the total surface area of the \(n\) smaller cubes is \(6n\text{ cm}^2\).
Since the total surface area of the \(n\) cubes is ten times the surface area of Caen’s original cube, we have \[6n=10(6x^2)\] Dividing both sides by 6, we have \[n=10x^2\] But \(n=x^3\), so this tells us that \[x^3=10x^2\] Since \(x>0\), we have \(x^2>0\). Dividing both sides by \(x^2\), we find that \(x=10\).
Therefore, the side length of Caen’s original cube was \(10\) cm.
Extension:
If the combined surface area of the \(n\) cubes with a side length of \(1\) cm was \(Q\) times the surface area of the original uncut cube, then the side length of the original uncut cube would have been \(Q\) cm. Can you see why?