Cubical Truncations

By Euler's formula for polyhedra, we know that $f-e+v=2$ , so $f+e+v=2+2e$ . The original cube has 12 edges, and taking away one small cube increases the edges by 9. Therefore: $f+e+v=2+2e =2+2(8\times 9+12) =2+168 =170$ .

As the cube has 12 edges and cubes are cut from each corner ,in each corner there will be 9 edges.So,number of edges is 9*8+12+84. f+v+2+e.therefore sum of number of faces and edges is 86. so.f+v+e=84+86=170

The cube has originally 6 faces, 12 edges, and 8 vertices. To solve for the face of the new polyhedron, count the number of the faces formed by cutting out a cube, we have 3 new faces, and multiply it by 8, since we have 8 cut out cubes. We now have 24 new faces plus the 6 existing face for a total of 30 faces.

Next, to solve for the edge of the new polyhedron, count the number of the new edges formed, we have 9 multiply by 8. Adding the existing and the new edges, we now have 84 edges.

Lastly, to solve for the vertices, we don't need to add the existing and the new vertices since two edges shared a vertex. All we need to do is count the vertices of one cut out part and multiply it by 8. We have 7 vertices for one cut out part, and multiplying it by 8 gives us 56 vertices.

We have now values for f, e, and v. These are 30, 84, and 56 respectively. Therefore, f+e+v gives us 170.

Euler's Formula for a polyhedron states that F + V - E = 2 We have to find F + V + E. As the cube is a polyhedron and the smaller cubes are removed in a regular fashion (no disorted transformations / reshaping), we simply have to find: 2 + 2E (substitute Euler's formula in the given statement).

So, all we have to do is count the number of edges on the polyhedron. A cube has 12 edges. The exposed surface of the cut area will have 12 - 3 = 9 edges (If these 3 edges are present, it will form a closed cube and no longer be exposed).

Therefore, we have 9 * 8 = 72 new edges for the cut cubes. Also, we have the original 12 edges which remain unchanged (although their length is smaller now).

Thus, total edges = 12 + 72 = 84 And F + V + E = 2 + 2E = 170 ... QED!

Since you were given a cube, you should know that it has 6 faces, 8 vertices, and 12 edges. Since smaller cubes of side length of 1 are cut out of the larger cube's corners, there are 8 smaller cubes. By doing that you are going to lose all 8 vertices of the larger cube. The "imprint" left from the cutout cubes is going to have 3 faces, 7 vertices, and 9 edges. How? Well focus on a single spot of that large cube. (I focused on the cutout portion.) What you see on the smaller cube, is what you subtract. Since a cube is being cutout of the larger one, you can only see 3 faces of the cube, because the other three are embedded in the larger cube. You can see that there is also one vertex (ignore the vertices that are touching the original square, because you want to subtract what you are losing from the larger square). You can also see the length, width and the height of the cube. The length width and the height are all edges of a cube, so you subtract 3. But from what? Well, remember that any cube has 6 faces, 8 vertices, and 12 edges, so you would subtract from what you saw from the smaller cube still embedded in the larger cube. 6-3=3 faces 8-1=7 vertices 12-3=9 edges Since there were eight cubes that you cutout, you multiply the "leftovers" by 8. Which results in 24 faces, 56 vertices, and 72 edges. And no, you don't add them and say your done, BECAUSE you still have the larger cube to worry about. So in the larger cube the 6 faces still remain in tact, even if you take a piece of it. The 8 vertices are all cutoff, because they were part of the smaller cubes that were cutout. The 12 edges still remain, also even if you take a portion of it away. The result of the leftovers of the larger cube are 6 faces, 0 vertices, and 12 edges. Now you would add the remains from the larger cube to the leftovers from smaller cubes. 24+6=30 faces 56+0=56 vertices 72+12=84 edges Then you add the faces, vertices and the edges together to get 170.

Large Cube Smaller Cube

F V E | F V E 6 8 12 | 6 8 12 -8 | -3 -1 -3

6 0 12 | 3 7 9 8 x ( 3 7 9 ) ^| 24 56 72 Then add the above to --->

F + V + E 30 + 56 + 72 = 170

* You can also use Euler's formula. Just find out two of the variables, then plug them into the equation F+V-E=2 to find out the third variable and then add all three.

* You can also do it the long way and take a Rubik's cube, rip off the corner cubes(if you want to) and count.

P.S. Is it possible to draw a diagram for this, in this solution box?

First Step:- 3 Extra faces for every corner is F= 6+3x8=30 Second Step:- 9 Extra edges for every corner is E= 12+9x8=84 Third Step:- Every vertex is replaced by 7 vertices V= 7x8=56 F+E+V= 30+84+8=170

Originally, the cube has $6$ faces, $12$ edges and $8$ vertices. At each vertex, by cutting out the cube, we will get $3$ additional faces, $9$ new edges, $7$ new vertices, but lose the corner vertex. Hence, $f = 6 + 3 \times 8 = 30$ , $e = 12 + 9 \times 8= 84$ and $v = 8 + 6 \times 8= 56$ . Thus, $f + v + e = 30 + 84 + 56 = 170$ .

Note: The Euler's characteristic for a polyhedron tells us that $f - e + v = 2$ . Hence, we need only calculate $e=84$ to arrive at the answer.