The figure below shows a small cube and a large cube. The length of each small cube is half the length of each large cube. Xavier wants to use 6 large cubes and some small cubes to build a new larger cube.

1. Express the volume of one large cube in terms of small cubes.

Since the length of a large cube is twice that of a small cube, its volume is $2 \times 2 \times 2 = 8$ times the volume of a small cube.
$1 \text{ large cube} = 8 \text{ small cubes}$

2. Find the total volume of the 6 large cubes.

Volume of 6 large cubes $= 6 \times 8 = 48 \text{ small cubes}$

3. Determine the volume of the smallest possible new larger cube.

The new larger cube must have a volume that is a perfect cube (in terms of small cubes) and greater than 48.
The perfect cubes are $1^3=1$, $2^3=8$, $3^3=27$, $4^3=64$, etc.
The smallest perfect cube greater than 48 is 64.

4. Calculate the number of small cubes needed.

Number of small cubes needed $= 64 - 48 = 16$

1. Find the side length of the new larger cube.

Let the side length of the new larger cube be $L$.
$L^3 = 2744$
$L = \sqrt[3]{2744} = 14 \text{ cm}$

2. Determine the length of each small cube.

From part (a), the new larger cube is formed by $2 \times 2 \times 2 = 8$ large cube spaces, meaning its side length is twice that of a large cube, or $4$ times that of a small cube.
Let the length of the small cube be $x$.
$4x = 14$
$x = 14 \div 4 = 3.5 \text{ cm}$