It is given that A,B,C and D are roots of the quartic equation X^4 - X +2 = 0. Determine, whether or not (AB+CD) is a root of the equation X^3 – 8X –1 = 0.
Let A,B,C,D be the roots of Q(x) = x^4 - c_3*x^3 + c_2*x^2 - c_1*x + c_0. Let W=AB+CD, Y=AC+BD, Z=AD+BC.
The following identities relate A,B,C,D,X,Y,Z:
W+Y+Z = AB+AC+AD+BC+BD+CD
WY+WZ+YZ = (A+B+C+D) * (ABC+ABD+ACD+BCD) - 4*(ABCD)
WYZ = (A^2+B^2+C^2+D^2) * (ABCD) + ((ABC)^2+(ABD)^2+(ACD)^2+(BCD)^2)
A^2+B^2+C^2+D^2 = (A+B+C+D)^2 - (AB+AC+AD+BC+BD+CD)
(ABC)^2+(ABD)^2+(ACD)^2+(BCD)^2 = (ABC+ABD+ACD+BCD)^2 - 2*(ABCD)*(AB+AC+AD+BC+BD+CD)
From Q(x):
A+B+C+D = c_3
AB+AC+AD+BC+BD+CD = c_2
ABC+ABD+ACD+BCD = c_1
ABCD = c_0
Then:
(ABC)^2+(ABD)^2+(ACD)^2+(BCD)^2 = c_1^2-2*c_0*c_2
A^2+B^2+C^2+D^2 = c_3^2-c_2
WYZ = (c_3^2-c_2) * (c_0) + (c_1^2-2*c_0*c_2) = c_3^2*c_0 - 3*c_0*c_2 + c_1^2
WY+WZ+YZ = c_3*c_1 - 4*c_0
W+Y+Z = c_2
Then W,Y,Z are the roots of R(x) = x^3 - c_2*x^2 + (c_3*c_1 - 4*c_0)x - (c_3^2*c_0 - 3*c_0*c_2 + c_1^2)
From the given equation Q(x) = x^4 - x + 2: c_3 = 0, c_2 = 0, c_1 = 1, c_0 = 2. Then R(x) = x^3 - 0*x^2 + (0*1-4*2)x - (0^2*2-3*2*0+1^2) = x^3 - 8x - 1. This is the cubic equation sought, so the final answer is yes, AB+CD is a root of the cubic.
General Solution
