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)).
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Submitted by Dustin
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Rating: 2.7500 (4 votes)
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Solution:
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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. |
