how to you solve a rubik's cube?
414837
The idea is to build up the corners, you want to get a 2x2x2 corner lined up and then work from there.\n\nHere's a very helpful link\nhttp://lar5.com/cube/
465982
It cannot end in the middle.\n\nLet's look at the cube.  \nIt's divided 3x3 to give you 27 subcubes.\nThe middle cube is therefore only one deep from any side.\nThere are eight subcubes on each corner of the cube.\nThere are six subcubes in the middle of each face of the cube.\nWe'll put all these subcubes into a group.\nCall it group A. 8+6=14 subcubes.\n\nAnd there are all the other cubes.\nThere are 27 subcubes.  14 are in group A.\nSo there are 27-14=13 of these.  \nWe'll group them together and call it group B.  13 subcubes.\n\nSome things become clear at this point.\nThe mouse could progress always ahead. So...\nEach time the mouse eats one of the group A subcubes, it must go to one of the group B subcubes.\n\nEach time it eats one of the group B subcubes, it must go to one of the group A subcubes.\n\nEven if it goes sideways, to a subcube in the same group (that's only possible for group B subcubes because they are the only subcubes where members of the group can be adjacent), what happens?\n\nThe number of subcubes in group B will reduce by two instead of one.\nThe number of subcubes in group A will remain the same as before.\n\nHowever the mouse moves and eats, there will always be more group A subcubes than there are group B subcubes.\n\nThe middle subcube belongs to group B.\nIt's one of the group of 13 subcubes.\n\nWhichever group the mouse eats from on the first day, he will always have one of group A subcubes left.\nBecause there are always more group A subcubes than group B subcubes.\n\nI'll repeat all that.\n27 subcubes, one per day = 27 days. \n13 subcubes in group B.\n14 subcubes in group A.\n\nIt is impossible for the mouse to have eaten the last group B subcube and there be no group A subcubes left.\nBecause there are always more group A subcubes than there are group B subcubes.\n\nSo the last subcube the mouse eats will always be one of group A.\nA corner subcube or a subcube that's in the middle of one of the faces.\nNot the subcube in the middle.\nQED, I think.
618016
4x4x4=64-4=60\n\nstill trying\n\nok, got it\n\n(4cubed) = 64\n4/4 =1\n64-1-4=59\n\n(4cubed)-4-(4/4)\n\nI cant get the small 3 sign for the cubed, so had to write it.\n\nReally pleased with myself, going for a beer to celebrate :)
240703
1^3 = 1^2\n4^3 = 8^2\n9^3 = 27^2\n\nsee the pattern?  cube a perfect square\n\nso 16^3 = 64^2 will work (4096)\n\n25^3 = 125^2  (you found)\n\n36^3 = 216^2  (46656)\n\ntoo high\n\nSo you only missed 4096.
504740
After several attempts, it seems that it is rather impossible.  If you start in the bottom corner, you can get all but 3 squares rather easily.  If you're careful, you can have 1 remaining if you start diagonal to the corner.\n\nWhat are you trying to do this for?  Is it a website?  I'd love to try it again.
485200
Practice!\n\nThis will help you learn to think in different ways that puzzles require, and to see patterns in the challenges presented by puzzles.
955733
http://software.rubikscube.info/
30747
2 cubes have 16 corners (since each cube has 8 corners)\n\na cylinder doesn't have any corners
396715
Took it last quarter, looked at it like a puzzle.  A puzzle is fun, math is not.  You start with the matrix given in the problem, and have to finish with ending matrix given.  Do what you need to do to get to the last matrix.  The best part, if you mess-up, just do another set.  There are no wrong answers, try doing it like a puzzle, more fun than Soduko.  Good Luck
992764
64.\n\n1 cubed = 1\n2 cubed =8\n3 cubed =27\n4 cubed =64\n5 cubed = 125 etc.
297409
This doesn't make any sense. You need to put down the numbers you're working with. If you're talking about finding cube roots, that's one thing; taking any number to the third power (x, for instance), is just: x*x*x.
277462
Within your 3x3x3 cube, we have ten classes of rectangular prisms, including the two boundary cases of the whole cube and of the single cublet:\n1x1x1, 1x1x2, 1x2x2, 1x1x3, 1x2x3, 2x2x2, 2x2x3, 1x3x3, 2x3x3, 3x3x3.\n\nFor each prism class, we have to count all possible arrangements within the main cube. The contributes to this number are from translations, then from rotations (but only for non-symmetric prisms). If I'm not wrong, they are:\n1x1x1: 27\n1x1x2: 54\n1x2x2: 12\n1x1x3: 27\n1x2x3: 18\n2x2x2: 8\n1x3x3: 9\n2x2x3: 12\n2x3x3: 6\n3x3x3: 1\n\nTotal=174 prisms\n\nAs per your later addition, if we want to discard cubic prisms, we have: 174-27-8-1 = 138 non-cubic prisms.
36132
http://www.websudoku.com/\n\nThere is a print option below the puzzle.
883909
draw FBDs of the 2 blocks independently\nbe careful\nfind the accelrations of the  2 blocks using CONSTRAINT relations\nand comes ur answer\n\ndon't wait for the solution\nit lessens ur thinking power n creativity\n\nif it's still problematic, consult ur lecturer\n\nconcentrate on ur studies \nur still a college guy
903120
Part One.  Let\nA = a sqrt(2) / 2 <i + j>  (magnitude a)\nB = a sqrt(2) <i + k>  (magnitude 2a)\nC = 3a sqrt(2) / 2 <j + k>  (magnitude 3a)\n\nThe resultant is\n\nR = a sqrt(2) <(3/2)i + 2j + (5/2)k>  (magnitude 5a)\n\nDirection cosines:\n\ncos alpha = 3 sqrt(2) / 10 = 0.424  (64.9 degrees)\ncos beta = 2 sqrt(2) / 5 = 0.566  (55.6 degrees)\ncos gamma = sqrt(2) / 2 = 0.707  (45 degrees)\n\n\nPart Two.  A = <i + j>; B = <i + k>; C = <j + k>\n\nDiagonal of cube D = <i + j + k>\n\nA + B + C = <2i + 2j + 2k> = 2D
869806
213\n321\n132\n\nall rows, columns, diagonals = 6\nnice try rt11guru but that wasn't your original answer\nmirror image:\n312\n123\n231\nupside down:\n231\n123\n312\nmirror image of that:\n132\n321\n213\n\nfuntym888 what do you mean it cant be done? dont you read the other answers before you answer?  it certainly can be done, in 4 different ways no less, as I just showed!  you simply take a 1-9 magic square and replace 4,5,6 by 1,2,3 and 7,8,9 by 1,2,3 respectively.  And it is "really" a magic square, it just isnt a normal magic square.  You should read your own sources as well!
695328
(1) add row#1*6 from row#2, then from row#3\nthat is [–6+1*6, –47+6*6, -2+1*6] for row#2\nthat is [-6+1*6, -41+6*6, 2+1*6] for row#3\nthus:\n1……6……1       \n0…..-11…..4\n0……-5…..8\n(2) add row#2*(-5/11) from row#3\nthat is [0, -5+(-5)*(-5/11), 8+4*(-5/11)] for row#3 \nthus:\n1……6……1       \n0…...-11…..4\n0……0….68/11\nto check the result multiply main diagonal: 1*(-11)*(68/11)=-68