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)).
(In reply to
answer by K Sengupta)
Comparing the coefficients of x^0, x^1, x^2 and x^3 in turn in the first equation, we have:
(i) 26D = 2197: (ii) AD = 26C; (iii) AC = 26B; (iv) AB = 8
From (i) and (iv), we have:
(D, B) = (2197/26, 8/A) = (169/2, 8/A)
Thus, AD = 26C gives:
A(169/2) = 26C
or, A = (4/13)*C
From (iii), we have:
(A^2)*C = 26*8 = 208
or, C^3 = (208*169)/16 = 13^3
or, C = 13, giving: A = (4/13)*13 = 4
Thus, B = 8/A = 8/4 = 2
Accordingly,
(13B/2 + 2D/13 + AB)
= 13*(2/2) + 2*{(2197/26)/13} + 2*4
= 34
For the second equation, an instant comparison of the coefficients of x^0 and x^3 gives:
F = 8/E, and: H = 169/4
The comparison of the coefficients of other exponents of x yield:
(EG, EH) = (52F, 52G)
or, (EG, E) = (416/E, (16/13)*G)
or, (E^2*G, E) = (416,(16/13)*G)
Thus, (16/13)^2*(G^3) = 336
or, G^3 = 2197/8, giving:
G = 13/2, so that:
E = (16/13)*(13/2) = 8, and:
F = 8/8 = 1
Accordingly,
13F/2 + 2H/13 + EF
= 13/2 + (2/13)*(169/4) + 8
= 21
Consequently, the value of the given expression is 34+21 = 55.
Puzzle Solution
