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# Cubes Within Cubes Revisited

##### Age 11 to 14Challenge Level

Yanqing from Devonport High School for Girls sent us a complete solution to this problem:

We would need to use $26$ red cubes and $98$ blue cubes.

For the $3 \times 3 \times 3$ cube, there is one yellow cube, so there are $27 - 1 = 26$ red cubes.

It is the same for the blue cubes: there are $5 \times 5 \times 5$ cubes in total, and $3 \times 3 \times 3$ of them are not blue.

For the next size up, there would be $7 \times 7 \times 7$ cubes in total, and $5 \times 5 \times 5$ that aren't green.

So there would be $343 - 125 = 218$ green cubes.

The amount you need for each layer added onto a cube with edges of $x$ will be $(x+2)^3 - x^3$.

$(x+2)^3$ is the cube you will get when you add the next layer, and $x^3$ is the number of cubes already there. (This is Martha's method)

With Emma's method, the top and bottom of the layer would both contain $(x+2)^2$ cubes.

Each added face has $x^2$ cubes, and the four vertical columns $4x$ cubes altogether.

So the expression for the total number of cubes needed would be $2(x+2)^2 + 4x^2 + 4x$.

With Charlie's method, the top and bottom of the layer would both contain $x^2$ cubes.

Each column has $x + 2$ cubes, and there would be $4(x + 1)$ columns.

So the expression for the total number of cubes needed would be $2x^2 + 4(x+1)(x+2) = 6x^2 + 12x + 8$.

Martha's method: \begin{eqnarray} (x+2)^3 - x^3 &=& x^3 + 6x^2 + 12x + 8 - x^3\\ &=&6x^2 + 12x + 8 \end{eqnarray} Emma's method: $$2(x+2)^2 + 4x^2 + 4x = 6x^2 + 12x + 8$$ Charlie's method: $$2x^2 + 4(x+1)(x+2) = 6x^2 + 12x + 8$$

These are all equivalent.

Another way of thinking about the problem is:

First you put a square with side of $x$ on each face ($6x^2$ cubes), then a column of $x$ cubes on each edge ($12x$ cubes), and finally one cube on each corner ($8$ cubes).

This would give us our $6x^2 + 12x + 8$ expression, which is what you get when you simplify the other expressions.

Well done Yanqing.