Rubo Cubo

For any cube with dimensions n n n and n>1, there will be a (n-2) (n-2) (n-2) cube inside it and just barely not reaching. The number of cubes visible is the area of the big cube - the area of the contained cube. In this case, it will be 7^3-5^3 = 218 cubes visible.

A cube has $6$ faces, $12$ edges, and $8$ vertices.

A face has $7\times 7=49$ cubes, so that will be total of $6\times49=294$ .

But the edges have been counted with both sides, so they need to be subtracted.

An edge has 7 cubes, so that is a total of $7\times12=84.$ Subtract that and we have $294-84=210$ .

But what about the vertices? They were each counted three times in the first round, and each was subtracted three times in the second, so they need to be counted. There are $8$ of them, each with only $1$ cube. So the final correction is $210+8=218$ .

According to the question, 2x2x2 = 8 pieces 3x3x3 = 26 pieces 4x4x4= 56 pieces If "n" is the amount of piece of a side of a cube (like n=3 for 3x3x3) Then, n=2, 2x2x2 = 8 = 2^3 - (n-2)^3 = 2^3 - (2-2)^3 n=3, 3x3x3 = 26 = 3^3 - (n-2)^3 = 3^3 - (3-2)^3 n=4, 4x4x4 = 56 = 4^3 - (n-2)^3 = 4^3 - (4-2)^3 So, for 7x7x7, n=7 7x7x7 = 7^3 - (n-2)^3 =7^3 - (7-2)^3 =7^3 - 5^3 =343 - 125 =218

Ans:218 pieces.