# resolvent cubic

#### apple2009

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 didn’t get the other two equal.

#### chiph588@

MHF Hall of Honor
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 didn’t get the other two equal.
Brute force computation my friend! This is the only way I know how to verify this...

I see your solving the quartic!

#### TheArtofSymmetry

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 didn’t get the other two equal.
You just need to plug Ѳ, Ѳ₂and Ѳ₃into the forms of elementary symmetric polynomials and verify that it indeed generates R(x).

See here and verify that e_1 = 2a, e_2= a^2-4c, e_3=-b^2

It requires a lengthy computation.

#### simplependulum

MHF Hall of Honor
You just need to plug Ѳ, Ѳ₂and Ѳ₃into the forms of elementary symmetric polynomials and verify that it indeed generates R(x).

See here and verify that e_1 = 2a, e_2= a^2-4c, e_3=-b^2

It requires a lengthy computation.
This computation is indeed an art of symmetry !!