INTRODUCTORY NOTE:
This problem was originally inspired by a Mathematical Olympiad problem. However, the problem body corresponding to this exercise is separate from that of the Olympiad problem.
SOLUTION TO THE PROBLEM:
# (AB +CD ) is a root of the equation X^3 – 8X –1 = 0.
EXPLANATION:
Let us suppose P = A+B ; Q = AB ; R = C+D and S = CD. Then, equating the coefficients of the various exponents of X in the equality X^4 - X +2 = (X-A)(X-B)(X-C)(X-D) ; we obtain :
I) 0 = A+B+C+D = P+R;
II) 0 = AB+AC+AD+BC+BD+CD =PR+Q+S ;
III) 1= ABC+ABD+ACD+BCD = QR+PS;
IV) 2 = ABCD = QS ;
From(I) , we obtain R =-P and substituting this in (II) we obtain Q+S = - PR = P^2-------(V). But (IV) gives Q = 2*(1/S) and, accordingly from (III) and (IV) we obtain
S + 2*(1/S) = P^2 and P { S – 2*(1/S)} =1 ;
or, S + 2*(1/S) = P^2 and S – 2*(1/S) = 1/P ;
or, 8 = P^4 – (1/P)^2 so that, P^6 – 8*(P^2) – 1 = 0....... (m)
But we recall from (V) that P^2 = Q+S = AB+CD.
# Consequently,, substituting X = P^2 in condition (m), we observe that, (AB +CD ) IS INDEED A ROOT OF THE EQUATION X^3 – 8X –1 = 0.
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