is there a formula to solve the rubik's cube?
96945
I answered a same question before.\n\nhttp://answers.yahoo.com/question/index;_ylt=ArTjPdUxvwSu88cv2FyCYxvsy6IX?qid=1005122402880
1180807
Yes its the Rubik's cube. I remember when it was popular. I hated it. There was even a TV special on 101 ways to destroy it. Maybe that is why I am a wee bit off.
532355
Please see the method explained by me as an answer tothis question..\nhttp://in.answers.yahoo.com/question/index;_ylt=AiNJtRa.xTUMUTdP29hR7sSQHQx.?qid=1006051102360\n\nThis will help u solve cubic as well as other equations.. :-)
1076564
What do you want to solve for? x? y? or do you want an ordered pair (x,y) that works (integers maybe?)
769753
Without knowing the relationship between successive elements, I think the best you can do is this:\n[k=0, n-1]∑(x_(k+1) - x_k)²\n[k=0, n-1]∑(x_(k+1)² - 2x_k*x_(k+1) + x_k²)\n[k=0, n-1]∑x_(k+1)² - 2[k=0, n-1]∑(x_k*x_(k+1)) + [k=0, n-1]∑x_k²\n[k=1, n]∑x_k² - 2[k=0, n-1]∑(x_k*x_(k+1)) + [k=0, n-1]∑x_k²\n\nx_0 + x_n + 2[k=1, n-1]∑x_k² - 2[k=0, n-1]∑(x_k*x_(k+1))
293873
http://www.geocities.com/jasmine_ellen/RubiksCubeSolution.html will get you under a minute with practice.
1136055
Here's how we can solve the problem logically, without doing all the multiplication.\n\nFirst of all, the coefficients aren't relevant to the problem; we don't care what the terms are specifically, just how many of them there are.\n\nLet's create two new temporary variables: let a = x + 5y, and b = 3z + w.  That reduces the problem, for the moment, to (a + b)^6.\n\nNow, any time you take a binomial to a power -- such as (a+b)^n -- there will be n+1 terms in the final expansion: (a+b)^2 has 3 terms, (a+b)^3 has 4 terms, etc.\n\nSo in (a+b)^6 there will be seven terms, which will have various coefficients but generally be of the forms a^6, a^5b, a^4b^2, a^3b^3, a^2b^2, ab^5, and b^6.  Note that, in each term, the sum of the powers is 6.\n\nNow, if we reinstate the original variables, substituting back in for a and b, we can consider each of those seven terms.\n\na^6 becomes (x+5y)^6, which will have 7 terms.\na^5b becomes (x+5y)^5 (3z+w), which will have 6 * 2 = 12 terms.\na^4b^2 becomes (x+5y)^4 (3z+w)^2, which will have 5 *3 = 15 terms.\na^3b^3 becomes (x+5y)^3 (3z+w)^3, which will have 4 * 4 = 16 terms.\na^2b^4 becomes (x+5y)^2 (3z+w)^4, which will have 3 *5 = 15 terms.\nab^5 becomes (x+5y) (3z+w)^5, which will have 2 *6 = 12 terms.\nAnd, finally, b^6 will have 7 terms.\n\n7+12+15+16+15+12+7 = 84.\n\nHope that made sense.
1169447
The Cube has 43,252,003,274,489,856,000 different possible configurations.\n\nI cheated and looked up the answer, but to figure it out you'll want to use combinatorics.  I'll have to work on that part of it...\n\nIf you're taking the cube apart and counting how many different ways you can put it back together, this is a relatively simple problem (although it doesn't match the above answer).\n\nThe center cubes don't move at all, so there is only one possible configuration of them.\n\nThere are 12 edge cubes, which each have 2 outside edges and therefore two orientations for each location.  The number of possible combinations of these edges is (2*12)*(2*11)*...*(2*2)*(2*1) = (2^12)*12!\n\nThere are 8 corner cubes, which each have 3 outside edges and therefore three orientations for each location.  The number of possible combinations of these corners is (3*8)*(3*7)*...*(3*2)*(3*1) = (3^8)*8!\n\nMultiply the number of possible combinations of edge cubes by the number of possible combinations of corner cubes and you end up with 519,024,039,293,878,300,000 combinations.  \n\nThis is different than the answer I looked up because there are possible combinations when you take the cube apart that cause it to no longer be solvable.  I can't think of an easy way to find the number of combinations that keep solvability.
471374
Dear student,\nThe cube root of 27225 is between 30.0831 and 30.0832.\nbest regards,\nMarcello
1187142
maybe something like 7^(1/3) over 7^(1/2)\n\nthat's the same as the cube root of 7 over the square root of 7.\nTo solve it, you would just subtract the exponents and you'd get 1 over 7^(1/6)\n\nI hope that is what you are looking for.
276464
(1x12)+(2x11)+(3x10)+(4x9)+\n(5x8)+(6x7)+(7x6)+(8x5)+\n(9x4)+(10x3)+(11x2)+(12x1)\n=12+22+30+36+40+42+42+\n40+36+30+22+12=364\n\n\n364 gifts
870731
A Rube Goldberg machine
1115504
the cube root can be found using a calculator\nmake sure you have a scientific one though\ni had do use it alot in geometry\n\nits alot like a square root just three instead of 2
133233
(1+1+1)! = 6 3 factorial is 6\n2+2+2 = 6\n3*3-3 = 6\nsquare root(4) + square root(4) + square root (4) = 6\n5 + 5/5 = 6\n6+6-6 = 6\n7 - 7/7 = 6\ncube root(8) + cube root(8) + cube root (8) = 6\n9+9 / square root(9) = 6
87438
cubed = power of 3\n\ny cubed = y raised to 3 or y^3 = y x y x y\n\n2 cubed = 2 ^ 3 = 2 x 2 x 2 = 8
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.
4519
(4/3)*pi*r^3\n\n[four thirds pi r cubed]\n\ngood luck!