PATTERNS
After filling in the table with the data I started looking for patterns so that I could work out formulas. I had to investigate and find formulas that would work out how many cubes had different amount of faces, e.g. 0 faces, 1 face, 2 faces, 3 faces.
I noticed that the columns for 3 faces had a pattern except 1 x 1 x 1. All of the cubes had 8 cubes with 3 different faces painted. All of these 8 are the vertices of the cube and so had three faces painted.
For 2 faces I noticed that each one got higher by 12, so it was + 12. This told me that somewhere in the formula that there would be +12.
For 1 face I knew that there would be a 6 in the formula, because all the numbers were multiples of 6. Whilst for 0 faces I could not see a pattern straight away, because there was a big leap for the numbers in the column for 0 faces.
The column for 4 and 5 faces was empty because on the large cube there can only be cubes with 0, 1, 2, 3 faces showing. However the cube with dimensions 1 x 1 x 1 had 6 faces showing which is an exception.
PREDICTION
FORMULAS
After making a prediction for a cube with the dimensions of 11 x 11 x 11, I then worked out formulas. Below are the formulas I found in this investigation, I explained how I worked out these formulas in my results. The letter n represents the length of a side.
0 Faces.
I knew that this formula would be cubed because cubes with 0 faces are inside the big cube and found that the third difference was constant, so I knew it would be a cubic equation. Looking at the numbers in the 0 faces column I found they were cube numbers, so I tried n3. This didn’t work so I looked closely and saw that the ‘n cubed’ was two numbers before it so I tried (n – 2). However this did work and so I found the formula (n – 2)3.
1 Face.
I found that the cubes with 1 painted face were on the middle of each face. There are 6 faces on a cube which made up the square. I also found that the
cubes with 1 face were all multiples of 6. It is like working out the area of the square so this why it was squared. So I got the formula 6(n-2)2.
2 Faces.
On a cube there are 12 edges altogether, so from this I knew I had to multiply something by 12.
I got the formula 12n-24.
3 Faces.
There are always 8 cubes with 3 faces; this is because there are eight vertices. This is not true for 1 x 1 x 1 because this is made up of 1 cube.
GENERAL FORMULA
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I then checked to see if the formulas worked because if you add them up then it should be equal to the total number of small cubes in the large cube. Which has the formula x=n3, (x representing the total number of cubes).
- With this it would also work out the general formula, so I added the formulas for 0 faces, 1 face, 2 faces and 3 faces.
Here are the formulas when added together:
= (n – 2)3 + 6(n – 2)2 + (12n +24) + (8)
= (n3 – 6n2 + 12n – 8) + (6n2 – 24n + 24) + (12n – 24) + (8)
= n3 – 6n2 + 12n – 8 + 6n2 – 24n + 24 + 12n – 24 + 8
= n3 + 12n – 8 -24n + 24 + 12n – 24 + 8
= n3 + 12n – 8 –24n + 24 + 12n – 24 + 8
= n3 – 8 +24 –24 + 8
= n3 – 8 +24 –24 + 8
= n3 +24 –24
= n3 +24 – 24
= n3
CONCLUSION
I found that the shape of the cube had a part in the formulas, like the number of cubes with painted faces was 8, because there are 8 vertices. Also on 2 faces 12 was to be multiplied by something because there are 12 edges.
