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 Difference of Cubes OF DOOM (Posted on 2005-02-11)
Suppose we have the follwing two identities:
(Ax-26)(Bx^2+Cx+D)=8x^3-2197
and (Ex-52)(Fx^2+Gx+H)=8x^3-2197;

Compute the quantity ((13B/2 + 2D/13 + AB)+(13F/2+2H/13+EF)).

 Submitted by Dustin Rating: 2.7500 (4 votes) Solution: (Hide) 55. np_rt has a much simpler method. Please read the comment list. If you multiply out (Ax - 26)(Bx^2 + Cx + D), you get ABx^3 + (AC - 26B)x^2 + (AD - 26C)x - 26D. Since this = 8x^3 - 2197, and the coefficients in front of each degree of x must be the same: AB = 8, AC - 26B = 0, AD - 26C = 0, and -26D = -2197. Therefore, D = 169/2. AC = 26B, and AD = 26C. Since A = 8/B and AD = 26C, 8D/B = 26C, so 4D = 13BC. Since D = 169/2, BC = 26, and C = 26/B. Since A = 8/B and AC = 26B, 8C/B = 26B, so 4C = 13B^2. Since 4C = 13B^2 and C = 26/B, 104/B = 13B^2, so 8 = B^3. B = 2. So C = 13. So A = 4. So, to answer the first part of the question: (13B/2 + 2D/13 + AB) = (13 + 13 + 8) = 34. If you multiply out (Ex - 52)(Fx^2 + Gx + H), you get EFx^3 + (EG - 52F)x^2 + (EH - 52G)x -52H. Again, the coefficients in front of degrees of x must be the same: EF = 8, EG - 52F = 0, EH - 52G = 0, and -52H = -2197. Therefore, H = 169/4, EG = 52F, and EH = 52G. Since E = 8/F and EH = 52G, 8H/F = 52G, so 2H = 13FG. Since H = 169/4, FG = 13/2, and 2G = 13/F. Since E = 8/F and EG = 52F, 8G/F = 52F, so 2G = 13F^2. Since 2G = 13F^2 and 2G = 13/F, 13/F = 13F^2, so 1 = F^3. F = 1. So G = 13/2, and E = 8. So, to answer the second part of the question: (13F/2 + 2H/13 + EF) = (13/2 + 13/2 + 8) = 21 Sum: 55.

 Subject Author Date Puzzle Solution K Sengupta 2008-07-07 15:21:12 answer K Sengupta 2008-07-05 12:25:57 Simpler Method np_rt 2005-02-12 18:20:57 re: No Subject -- NO ERROR Richard 2005-02-11 19:18:00 re(3): No Subject -- No Error Richard 2005-02-11 18:57:38 re(2): No Subject -- No Error Dustin 2005-02-11 17:41:54 re: No Subject -- Error? Richard 2005-02-11 17:34:55 No Subject Jun 2005-02-11 15:46:39

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