Color the cubes, then assemble them to form a larger cube












11














Goal: Paint 27 cubes using three colors (for example, red, yellow, and blue), so that you can form a 3x3x3 cube with all surfaces in red (for example), a 3x3x3 cube all in yellow, and a 3x3x3 cube all in blue by rearranging the painted cubes, not repainting them.










share|improve this question









New contributor




Daniel Mathias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • Welcome to PSE! Your question seems unclear to me. Why not just put the cubes into a 3x3x3 cube shape, paint the faces of the larger cube, disassemble, and then paint the rest of the faces whatever colors we want?
    – Frpzzd
    Dec 29 '18 at 22:06








  • 1




    You must be able to assemble the cube in each of the colors.
    – Daniel Mathias
    Dec 29 '18 at 22:09










  • Oh, I see. Thanks for clarifying!
    – Frpzzd
    Dec 29 '18 at 22:10






  • 1




    I ran into the same problem as @Frpzzd, so I tried to edit the question to be a bit more difficult to misunderstand. My sentence structure still seems a bit convoluted, so please feel free to improve it however you see fit.
    – Bass
    Dec 30 '18 at 0:22
















11














Goal: Paint 27 cubes using three colors (for example, red, yellow, and blue), so that you can form a 3x3x3 cube with all surfaces in red (for example), a 3x3x3 cube all in yellow, and a 3x3x3 cube all in blue by rearranging the painted cubes, not repainting them.










share|improve this question









New contributor




Daniel Mathias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • Welcome to PSE! Your question seems unclear to me. Why not just put the cubes into a 3x3x3 cube shape, paint the faces of the larger cube, disassemble, and then paint the rest of the faces whatever colors we want?
    – Frpzzd
    Dec 29 '18 at 22:06








  • 1




    You must be able to assemble the cube in each of the colors.
    – Daniel Mathias
    Dec 29 '18 at 22:09










  • Oh, I see. Thanks for clarifying!
    – Frpzzd
    Dec 29 '18 at 22:10






  • 1




    I ran into the same problem as @Frpzzd, so I tried to edit the question to be a bit more difficult to misunderstand. My sentence structure still seems a bit convoluted, so please feel free to improve it however you see fit.
    – Bass
    Dec 30 '18 at 0:22














11












11








11


0





Goal: Paint 27 cubes using three colors (for example, red, yellow, and blue), so that you can form a 3x3x3 cube with all surfaces in red (for example), a 3x3x3 cube all in yellow, and a 3x3x3 cube all in blue by rearranging the painted cubes, not repainting them.










share|improve this question









New contributor




Daniel Mathias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











Goal: Paint 27 cubes using three colors (for example, red, yellow, and blue), so that you can form a 3x3x3 cube with all surfaces in red (for example), a 3x3x3 cube all in yellow, and a 3x3x3 cube all in blue by rearranging the painted cubes, not repainting them.







geometry three-dimensional






share|improve this question









New contributor




Daniel Mathias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




Daniel Mathias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited Dec 30 '18 at 1:19









Brandon_J

1719




1719






New contributor




Daniel Mathias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked Dec 29 '18 at 21:57









Daniel Mathias

584




584




New contributor




Daniel Mathias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Daniel Mathias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Daniel Mathias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • Welcome to PSE! Your question seems unclear to me. Why not just put the cubes into a 3x3x3 cube shape, paint the faces of the larger cube, disassemble, and then paint the rest of the faces whatever colors we want?
    – Frpzzd
    Dec 29 '18 at 22:06








  • 1




    You must be able to assemble the cube in each of the colors.
    – Daniel Mathias
    Dec 29 '18 at 22:09










  • Oh, I see. Thanks for clarifying!
    – Frpzzd
    Dec 29 '18 at 22:10






  • 1




    I ran into the same problem as @Frpzzd, so I tried to edit the question to be a bit more difficult to misunderstand. My sentence structure still seems a bit convoluted, so please feel free to improve it however you see fit.
    – Bass
    Dec 30 '18 at 0:22


















  • Welcome to PSE! Your question seems unclear to me. Why not just put the cubes into a 3x3x3 cube shape, paint the faces of the larger cube, disassemble, and then paint the rest of the faces whatever colors we want?
    – Frpzzd
    Dec 29 '18 at 22:06








  • 1




    You must be able to assemble the cube in each of the colors.
    – Daniel Mathias
    Dec 29 '18 at 22:09










  • Oh, I see. Thanks for clarifying!
    – Frpzzd
    Dec 29 '18 at 22:10






  • 1




    I ran into the same problem as @Frpzzd, so I tried to edit the question to be a bit more difficult to misunderstand. My sentence structure still seems a bit convoluted, so please feel free to improve it however you see fit.
    – Bass
    Dec 30 '18 at 0:22
















Welcome to PSE! Your question seems unclear to me. Why not just put the cubes into a 3x3x3 cube shape, paint the faces of the larger cube, disassemble, and then paint the rest of the faces whatever colors we want?
– Frpzzd
Dec 29 '18 at 22:06






Welcome to PSE! Your question seems unclear to me. Why not just put the cubes into a 3x3x3 cube shape, paint the faces of the larger cube, disassemble, and then paint the rest of the faces whatever colors we want?
– Frpzzd
Dec 29 '18 at 22:06






1




1




You must be able to assemble the cube in each of the colors.
– Daniel Mathias
Dec 29 '18 at 22:09




You must be able to assemble the cube in each of the colors.
– Daniel Mathias
Dec 29 '18 at 22:09












Oh, I see. Thanks for clarifying!
– Frpzzd
Dec 29 '18 at 22:10




Oh, I see. Thanks for clarifying!
– Frpzzd
Dec 29 '18 at 22:10




1




1




I ran into the same problem as @Frpzzd, so I tried to edit the question to be a bit more difficult to misunderstand. My sentence structure still seems a bit convoluted, so please feel free to improve it however you see fit.
– Bass
Dec 30 '18 at 0:22




I ran into the same problem as @Frpzzd, so I tried to edit the question to be a bit more difficult to misunderstand. My sentence structure still seems a bit convoluted, so please feel free to improve it however you see fit.
– Bass
Dec 30 '18 at 0:22










4 Answers
4






active

oldest

votes


















13














Paint the small cubes with the colors red, blue, and green as follows, so that on each cube, the faces with the same color are all mutually adjacent.




  • One cube with three red faces and three blue faces

  • One cube with three red faces and three green faces

  • One cube with three blue faces and three green faces

  • Three cubes each with three red faces, two blue faces, and one green face

  • Three cubes each with three red faces, two green faces, and one blue face

  • Three cubes each with three blue faces, two red faces, and one green face

  • Three cubes each with three blue faces, two green faces, and one red face

  • Three cubes each with three green faces, two red faces, and one blue face

  • Three cubes each with three green faces, two blue faces, and one red face

  • Six cubes with two faces of each color


Then you have exactly the correct number of corner, edge, face-center, and center pieces for each cube.






share|improve this answer





















  • Much better than my answer; well done!
    – Hugh
    Dec 29 '18 at 23:12










  • @Hugh Thanks! It would be nice to generalize it. The 4x4x4 cube with 3 colors is pretty much trivial, but with 4 colors... hmm...
    – Frpzzd
    Dec 29 '18 at 23:22










  • Oh, that's a good puzzle... A 4*4*4 with 3 colours can't be too difficult, but with 4 colours it could get tough. I'm pretty sure it could be done though! You know what I'll be working on ;)
    – Hugh
    Dec 29 '18 at 23:33






  • 7




    This looks very like the expansion of $(RG+GB+BR)^3$ - I haven't figured out quite how yet though!
    – JonMark Perry
    Dec 30 '18 at 0:27






  • 1




    @Hugh The 4*4*4 with 4 colors is has a similar solution
    – Daniel Mathias
    Dec 30 '18 at 3:56



















8














Here is a simple painting procedure that leads to a solution for any size of cube.




  1. Start with $27$ (or $64$, or $n^3$) unpainted cubes assembled into a large cube.

  2. Paint the outside faces one colour.

  3. Rearrange the cubes by removing the bottom layer and placing it on top. Similarly take the left-most slice and place it against the right hand side. Finally take the rear slice and place that against the front face. This puts unpainted surfaces on the outside of the cube.

  4. Repeat steps 2 and 3 using different colours of paint until all cubes are fully painted.


This method works for any size of cube. Note also that in the finished product, every pair of adjacent cubes has touching faces that share the same colour.






share|improve this answer





















  • Beautiful stuff
    – Jonathan Allan
    Dec 30 '18 at 19:46



















2














If we use colours R, G and B, then we can write:



$$(RG+GB+BR)^3$$



which expands into



$$R^3G^3+G^3B^3+B^3R^3+3(R^2G^3B+G^2B^3R+B^2R^3G+R^3G^2B+R^2GB^3+RG^3B^2)+6R^2G^2B^2$$



If we let, w.l.o.g., $R=x, G=B=1$, then we have:



$$(2x+1)^3=8x^3+12x^2+6x+1$$



This describes the number of corner pieces, edge pieces and centre pieces, and the coefficients total $27$.



To explain a motive, we can see that each colour needs $54$ faces, and this fits exactly into the $27times6$ faces we have available.



Therefore the non-visible central cube cannot have any of the colours of the visible colour chose. Therefore there must be three of these, and also as we cannot waste space, they must consist of $2$ sets of $3$ colours. We now need to find six more corners for each colour - these cannot be on the same cube for the same reasons as above, and so we have partially coloured $3times6=18$ more cubes, leaving $6$. These six must have $2$ of each colour, again for the same reasons. The last pieces then fall into place.



Now we can attempt a $4times4times4$ colouring. The 'characteristic polynomial' we are looking for is:



$$8x^3+24x^2+24x+8=(2x+2)^3$$



so we need $4$ (symmetric) elements of two each.



I propose:



$$(RB+GB+BY+YR)^3$$






share|improve this answer





























    1














    This sounds very similar to...




    this puzzle featured in a video by TED-ED. I'll try and make a picture when I get home.







    share|improve this answer





















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      4 Answers
      4






      active

      oldest

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      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      13














      Paint the small cubes with the colors red, blue, and green as follows, so that on each cube, the faces with the same color are all mutually adjacent.




      • One cube with three red faces and three blue faces

      • One cube with three red faces and three green faces

      • One cube with three blue faces and three green faces

      • Three cubes each with three red faces, two blue faces, and one green face

      • Three cubes each with three red faces, two green faces, and one blue face

      • Three cubes each with three blue faces, two red faces, and one green face

      • Three cubes each with three blue faces, two green faces, and one red face

      • Three cubes each with three green faces, two red faces, and one blue face

      • Three cubes each with three green faces, two blue faces, and one red face

      • Six cubes with two faces of each color


      Then you have exactly the correct number of corner, edge, face-center, and center pieces for each cube.






      share|improve this answer





















      • Much better than my answer; well done!
        – Hugh
        Dec 29 '18 at 23:12










      • @Hugh Thanks! It would be nice to generalize it. The 4x4x4 cube with 3 colors is pretty much trivial, but with 4 colors... hmm...
        – Frpzzd
        Dec 29 '18 at 23:22










      • Oh, that's a good puzzle... A 4*4*4 with 3 colours can't be too difficult, but with 4 colours it could get tough. I'm pretty sure it could be done though! You know what I'll be working on ;)
        – Hugh
        Dec 29 '18 at 23:33






      • 7




        This looks very like the expansion of $(RG+GB+BR)^3$ - I haven't figured out quite how yet though!
        – JonMark Perry
        Dec 30 '18 at 0:27






      • 1




        @Hugh The 4*4*4 with 4 colors is has a similar solution
        – Daniel Mathias
        Dec 30 '18 at 3:56
















      13














      Paint the small cubes with the colors red, blue, and green as follows, so that on each cube, the faces with the same color are all mutually adjacent.




      • One cube with three red faces and three blue faces

      • One cube with three red faces and three green faces

      • One cube with three blue faces and three green faces

      • Three cubes each with three red faces, two blue faces, and one green face

      • Three cubes each with three red faces, two green faces, and one blue face

      • Three cubes each with three blue faces, two red faces, and one green face

      • Three cubes each with three blue faces, two green faces, and one red face

      • Three cubes each with three green faces, two red faces, and one blue face

      • Three cubes each with three green faces, two blue faces, and one red face

      • Six cubes with two faces of each color


      Then you have exactly the correct number of corner, edge, face-center, and center pieces for each cube.






      share|improve this answer





















      • Much better than my answer; well done!
        – Hugh
        Dec 29 '18 at 23:12










      • @Hugh Thanks! It would be nice to generalize it. The 4x4x4 cube with 3 colors is pretty much trivial, but with 4 colors... hmm...
        – Frpzzd
        Dec 29 '18 at 23:22










      • Oh, that's a good puzzle... A 4*4*4 with 3 colours can't be too difficult, but with 4 colours it could get tough. I'm pretty sure it could be done though! You know what I'll be working on ;)
        – Hugh
        Dec 29 '18 at 23:33






      • 7




        This looks very like the expansion of $(RG+GB+BR)^3$ - I haven't figured out quite how yet though!
        – JonMark Perry
        Dec 30 '18 at 0:27






      • 1




        @Hugh The 4*4*4 with 4 colors is has a similar solution
        – Daniel Mathias
        Dec 30 '18 at 3:56














      13












      13








      13






      Paint the small cubes with the colors red, blue, and green as follows, so that on each cube, the faces with the same color are all mutually adjacent.




      • One cube with three red faces and three blue faces

      • One cube with three red faces and three green faces

      • One cube with three blue faces and three green faces

      • Three cubes each with three red faces, two blue faces, and one green face

      • Three cubes each with three red faces, two green faces, and one blue face

      • Three cubes each with three blue faces, two red faces, and one green face

      • Three cubes each with three blue faces, two green faces, and one red face

      • Three cubes each with three green faces, two red faces, and one blue face

      • Three cubes each with three green faces, two blue faces, and one red face

      • Six cubes with two faces of each color


      Then you have exactly the correct number of corner, edge, face-center, and center pieces for each cube.






      share|improve this answer












      Paint the small cubes with the colors red, blue, and green as follows, so that on each cube, the faces with the same color are all mutually adjacent.




      • One cube with three red faces and three blue faces

      • One cube with three red faces and three green faces

      • One cube with three blue faces and three green faces

      • Three cubes each with three red faces, two blue faces, and one green face

      • Three cubes each with three red faces, two green faces, and one blue face

      • Three cubes each with three blue faces, two red faces, and one green face

      • Three cubes each with three blue faces, two green faces, and one red face

      • Three cubes each with three green faces, two red faces, and one blue face

      • Three cubes each with three green faces, two blue faces, and one red face

      • Six cubes with two faces of each color


      Then you have exactly the correct number of corner, edge, face-center, and center pieces for each cube.







      share|improve this answer












      share|improve this answer



      share|improve this answer










      answered Dec 29 '18 at 22:57









      Frpzzd

      871120




      871120












      • Much better than my answer; well done!
        – Hugh
        Dec 29 '18 at 23:12










      • @Hugh Thanks! It would be nice to generalize it. The 4x4x4 cube with 3 colors is pretty much trivial, but with 4 colors... hmm...
        – Frpzzd
        Dec 29 '18 at 23:22










      • Oh, that's a good puzzle... A 4*4*4 with 3 colours can't be too difficult, but with 4 colours it could get tough. I'm pretty sure it could be done though! You know what I'll be working on ;)
        – Hugh
        Dec 29 '18 at 23:33






      • 7




        This looks very like the expansion of $(RG+GB+BR)^3$ - I haven't figured out quite how yet though!
        – JonMark Perry
        Dec 30 '18 at 0:27






      • 1




        @Hugh The 4*4*4 with 4 colors is has a similar solution
        – Daniel Mathias
        Dec 30 '18 at 3:56


















      • Much better than my answer; well done!
        – Hugh
        Dec 29 '18 at 23:12










      • @Hugh Thanks! It would be nice to generalize it. The 4x4x4 cube with 3 colors is pretty much trivial, but with 4 colors... hmm...
        – Frpzzd
        Dec 29 '18 at 23:22










      • Oh, that's a good puzzle... A 4*4*4 with 3 colours can't be too difficult, but with 4 colours it could get tough. I'm pretty sure it could be done though! You know what I'll be working on ;)
        – Hugh
        Dec 29 '18 at 23:33






      • 7




        This looks very like the expansion of $(RG+GB+BR)^3$ - I haven't figured out quite how yet though!
        – JonMark Perry
        Dec 30 '18 at 0:27






      • 1




        @Hugh The 4*4*4 with 4 colors is has a similar solution
        – Daniel Mathias
        Dec 30 '18 at 3:56
















      Much better than my answer; well done!
      – Hugh
      Dec 29 '18 at 23:12




      Much better than my answer; well done!
      – Hugh
      Dec 29 '18 at 23:12












      @Hugh Thanks! It would be nice to generalize it. The 4x4x4 cube with 3 colors is pretty much trivial, but with 4 colors... hmm...
      – Frpzzd
      Dec 29 '18 at 23:22




      @Hugh Thanks! It would be nice to generalize it. The 4x4x4 cube with 3 colors is pretty much trivial, but with 4 colors... hmm...
      – Frpzzd
      Dec 29 '18 at 23:22












      Oh, that's a good puzzle... A 4*4*4 with 3 colours can't be too difficult, but with 4 colours it could get tough. I'm pretty sure it could be done though! You know what I'll be working on ;)
      – Hugh
      Dec 29 '18 at 23:33




      Oh, that's a good puzzle... A 4*4*4 with 3 colours can't be too difficult, but with 4 colours it could get tough. I'm pretty sure it could be done though! You know what I'll be working on ;)
      – Hugh
      Dec 29 '18 at 23:33




      7




      7




      This looks very like the expansion of $(RG+GB+BR)^3$ - I haven't figured out quite how yet though!
      – JonMark Perry
      Dec 30 '18 at 0:27




      This looks very like the expansion of $(RG+GB+BR)^3$ - I haven't figured out quite how yet though!
      – JonMark Perry
      Dec 30 '18 at 0:27




      1




      1




      @Hugh The 4*4*4 with 4 colors is has a similar solution
      – Daniel Mathias
      Dec 30 '18 at 3:56




      @Hugh The 4*4*4 with 4 colors is has a similar solution
      – Daniel Mathias
      Dec 30 '18 at 3:56











      8














      Here is a simple painting procedure that leads to a solution for any size of cube.




      1. Start with $27$ (or $64$, or $n^3$) unpainted cubes assembled into a large cube.

      2. Paint the outside faces one colour.

      3. Rearrange the cubes by removing the bottom layer and placing it on top. Similarly take the left-most slice and place it against the right hand side. Finally take the rear slice and place that against the front face. This puts unpainted surfaces on the outside of the cube.

      4. Repeat steps 2 and 3 using different colours of paint until all cubes are fully painted.


      This method works for any size of cube. Note also that in the finished product, every pair of adjacent cubes has touching faces that share the same colour.






      share|improve this answer





















      • Beautiful stuff
        – Jonathan Allan
        Dec 30 '18 at 19:46
















      8














      Here is a simple painting procedure that leads to a solution for any size of cube.




      1. Start with $27$ (or $64$, or $n^3$) unpainted cubes assembled into a large cube.

      2. Paint the outside faces one colour.

      3. Rearrange the cubes by removing the bottom layer and placing it on top. Similarly take the left-most slice and place it against the right hand side. Finally take the rear slice and place that against the front face. This puts unpainted surfaces on the outside of the cube.

      4. Repeat steps 2 and 3 using different colours of paint until all cubes are fully painted.


      This method works for any size of cube. Note also that in the finished product, every pair of adjacent cubes has touching faces that share the same colour.






      share|improve this answer





















      • Beautiful stuff
        – Jonathan Allan
        Dec 30 '18 at 19:46














      8












      8








      8






      Here is a simple painting procedure that leads to a solution for any size of cube.




      1. Start with $27$ (or $64$, or $n^3$) unpainted cubes assembled into a large cube.

      2. Paint the outside faces one colour.

      3. Rearrange the cubes by removing the bottom layer and placing it on top. Similarly take the left-most slice and place it against the right hand side. Finally take the rear slice and place that against the front face. This puts unpainted surfaces on the outside of the cube.

      4. Repeat steps 2 and 3 using different colours of paint until all cubes are fully painted.


      This method works for any size of cube. Note also that in the finished product, every pair of adjacent cubes has touching faces that share the same colour.






      share|improve this answer












      Here is a simple painting procedure that leads to a solution for any size of cube.




      1. Start with $27$ (or $64$, or $n^3$) unpainted cubes assembled into a large cube.

      2. Paint the outside faces one colour.

      3. Rearrange the cubes by removing the bottom layer and placing it on top. Similarly take the left-most slice and place it against the right hand side. Finally take the rear slice and place that against the front face. This puts unpainted surfaces on the outside of the cube.

      4. Repeat steps 2 and 3 using different colours of paint until all cubes are fully painted.


      This method works for any size of cube. Note also that in the finished product, every pair of adjacent cubes has touching faces that share the same colour.







      share|improve this answer












      share|improve this answer



      share|improve this answer










      answered Dec 30 '18 at 4:40









      Jaap Scherphuis

      14.8k12565




      14.8k12565












      • Beautiful stuff
        – Jonathan Allan
        Dec 30 '18 at 19:46


















      • Beautiful stuff
        – Jonathan Allan
        Dec 30 '18 at 19:46
















      Beautiful stuff
      – Jonathan Allan
      Dec 30 '18 at 19:46




      Beautiful stuff
      – Jonathan Allan
      Dec 30 '18 at 19:46











      2














      If we use colours R, G and B, then we can write:



      $$(RG+GB+BR)^3$$



      which expands into



      $$R^3G^3+G^3B^3+B^3R^3+3(R^2G^3B+G^2B^3R+B^2R^3G+R^3G^2B+R^2GB^3+RG^3B^2)+6R^2G^2B^2$$



      If we let, w.l.o.g., $R=x, G=B=1$, then we have:



      $$(2x+1)^3=8x^3+12x^2+6x+1$$



      This describes the number of corner pieces, edge pieces and centre pieces, and the coefficients total $27$.



      To explain a motive, we can see that each colour needs $54$ faces, and this fits exactly into the $27times6$ faces we have available.



      Therefore the non-visible central cube cannot have any of the colours of the visible colour chose. Therefore there must be three of these, and also as we cannot waste space, they must consist of $2$ sets of $3$ colours. We now need to find six more corners for each colour - these cannot be on the same cube for the same reasons as above, and so we have partially coloured $3times6=18$ more cubes, leaving $6$. These six must have $2$ of each colour, again for the same reasons. The last pieces then fall into place.



      Now we can attempt a $4times4times4$ colouring. The 'characteristic polynomial' we are looking for is:



      $$8x^3+24x^2+24x+8=(2x+2)^3$$



      so we need $4$ (symmetric) elements of two each.



      I propose:



      $$(RB+GB+BY+YR)^3$$






      share|improve this answer


























        2














        If we use colours R, G and B, then we can write:



        $$(RG+GB+BR)^3$$



        which expands into



        $$R^3G^3+G^3B^3+B^3R^3+3(R^2G^3B+G^2B^3R+B^2R^3G+R^3G^2B+R^2GB^3+RG^3B^2)+6R^2G^2B^2$$



        If we let, w.l.o.g., $R=x, G=B=1$, then we have:



        $$(2x+1)^3=8x^3+12x^2+6x+1$$



        This describes the number of corner pieces, edge pieces and centre pieces, and the coefficients total $27$.



        To explain a motive, we can see that each colour needs $54$ faces, and this fits exactly into the $27times6$ faces we have available.



        Therefore the non-visible central cube cannot have any of the colours of the visible colour chose. Therefore there must be three of these, and also as we cannot waste space, they must consist of $2$ sets of $3$ colours. We now need to find six more corners for each colour - these cannot be on the same cube for the same reasons as above, and so we have partially coloured $3times6=18$ more cubes, leaving $6$. These six must have $2$ of each colour, again for the same reasons. The last pieces then fall into place.



        Now we can attempt a $4times4times4$ colouring. The 'characteristic polynomial' we are looking for is:



        $$8x^3+24x^2+24x+8=(2x+2)^3$$



        so we need $4$ (symmetric) elements of two each.



        I propose:



        $$(RB+GB+BY+YR)^3$$






        share|improve this answer
























          2












          2








          2






          If we use colours R, G and B, then we can write:



          $$(RG+GB+BR)^3$$



          which expands into



          $$R^3G^3+G^3B^3+B^3R^3+3(R^2G^3B+G^2B^3R+B^2R^3G+R^3G^2B+R^2GB^3+RG^3B^2)+6R^2G^2B^2$$



          If we let, w.l.o.g., $R=x, G=B=1$, then we have:



          $$(2x+1)^3=8x^3+12x^2+6x+1$$



          This describes the number of corner pieces, edge pieces and centre pieces, and the coefficients total $27$.



          To explain a motive, we can see that each colour needs $54$ faces, and this fits exactly into the $27times6$ faces we have available.



          Therefore the non-visible central cube cannot have any of the colours of the visible colour chose. Therefore there must be three of these, and also as we cannot waste space, they must consist of $2$ sets of $3$ colours. We now need to find six more corners for each colour - these cannot be on the same cube for the same reasons as above, and so we have partially coloured $3times6=18$ more cubes, leaving $6$. These six must have $2$ of each colour, again for the same reasons. The last pieces then fall into place.



          Now we can attempt a $4times4times4$ colouring. The 'characteristic polynomial' we are looking for is:



          $$8x^3+24x^2+24x+8=(2x+2)^3$$



          so we need $4$ (symmetric) elements of two each.



          I propose:



          $$(RB+GB+BY+YR)^3$$






          share|improve this answer












          If we use colours R, G and B, then we can write:



          $$(RG+GB+BR)^3$$



          which expands into



          $$R^3G^3+G^3B^3+B^3R^3+3(R^2G^3B+G^2B^3R+B^2R^3G+R^3G^2B+R^2GB^3+RG^3B^2)+6R^2G^2B^2$$



          If we let, w.l.o.g., $R=x, G=B=1$, then we have:



          $$(2x+1)^3=8x^3+12x^2+6x+1$$



          This describes the number of corner pieces, edge pieces and centre pieces, and the coefficients total $27$.



          To explain a motive, we can see that each colour needs $54$ faces, and this fits exactly into the $27times6$ faces we have available.



          Therefore the non-visible central cube cannot have any of the colours of the visible colour chose. Therefore there must be three of these, and also as we cannot waste space, they must consist of $2$ sets of $3$ colours. We now need to find six more corners for each colour - these cannot be on the same cube for the same reasons as above, and so we have partially coloured $3times6=18$ more cubes, leaving $6$. These six must have $2$ of each colour, again for the same reasons. The last pieces then fall into place.



          Now we can attempt a $4times4times4$ colouring. The 'characteristic polynomial' we are looking for is:



          $$8x^3+24x^2+24x+8=(2x+2)^3$$



          so we need $4$ (symmetric) elements of two each.



          I propose:



          $$(RB+GB+BY+YR)^3$$







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Dec 30 '18 at 11:33









          JonMark Perry

          17.5k63584




          17.5k63584























              1














              This sounds very similar to...




              this puzzle featured in a video by TED-ED. I'll try and make a picture when I get home.







              share|improve this answer


























                1














                This sounds very similar to...




                this puzzle featured in a video by TED-ED. I'll try and make a picture when I get home.







                share|improve this answer
























                  1












                  1








                  1






                  This sounds very similar to...




                  this puzzle featured in a video by TED-ED. I'll try and make a picture when I get home.







                  share|improve this answer












                  This sounds very similar to...




                  this puzzle featured in a video by TED-ED. I'll try and make a picture when I get home.








                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Dec 29 '18 at 22:42









                  Hugh

                  1,4581617




                  1,4581617






















                      Daniel Mathias is a new contributor. Be nice, and check out our Code of Conduct.










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                      Daniel Mathias is a new contributor. Be nice, and check out our Code of Conduct.












                      Daniel Mathias is a new contributor. Be nice, and check out our Code of Conduct.
















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