I see your solving the quartic!
Let f(x) =x^4+ax^2+bx+c, and r₁, r₂, r₃, r₄ be the roots of f(x). Consider the elements
Ѳ₁= (r₁+r₂) (r₃+r₄)
Ѳ₂= (r₁+r₃) (r₂+r₄)
Ѳ₃= (r₁+r₄) (r₂+r₃)
Show that Ѳ₁, Ѳ₂, Ѳ₃ are the roots of R(x) =x^3-2ax^2+ (a^2-4c) x+b^2, called the resolvent cubic of f(x).
I try to show that Ѳ₁+ Ѳ₂+ Ѳ₃=2a, Ѳ₁ Ѳ₂+ Ѳ₁Ѳ₃+Ѳ₂Ѳ₃=a^2-4c; Ѳ₁Ѳ₂ Ѳ₃=-b^2, after the computation I only find Ѳ₁+ Ѳ₂+ Ѳ₃=2a, and didnít get the other two equal.