Solving Cubic Equations with Lagrange Resolvent?

Question

I'm having difficulties understanding my textbook's decription of solving cubic equations using Lagrange Resolvents and symmetric polynomials.

Here's what I understand: $$ x^3 + px - q = (x-r)(x-s)(x-t)$$ We can also write: $$\lambda = r+ws+w^2t$$ $$\mu =wr+s+w^2t$$ where $1, w, w^2$ are the cubic roots of 1. I then understand that $\lambda^3 + \mu^3$ and $\lambda^3\mu^3$ are symmetric polynomials in r, s, and t. It is also solvable that the elemntary symmetric functions in r, s, t are $0, p, q$ where $$r+s+t=0$$ $$rs+rt+st=p$$ $$rst=q$$ The part where I get confused is that the book claims that $\lambda^3$ and $\mu^3$ are the roots of the quadratic polynomial $q(x)=x^2-(\lambda^3+\mu^3)x+\lambda^3\mu^3$, which seems obvious to me. Then they claim you can use the quadratic formula to solve for $\lambda^3$ and $\mu^3$ in terms of $p$ and $q$, thus allowing you to solve a system of equations to acquire $r,s,t$.

How can you use the quadratic formula to "explicitly solve for $\lambda^3$ and $\mu^3$ in terms of $p$ and $q$"?

Answer 1

A symmetric polynomial in $r$, $s$, and $t$ will be a polynomial in the elementary symmetric functions $0$, $p$, and $q$, i.e. a polynomial in p and q. You need to to know those polynomials explicitly to solve the quadratic equation explicitly. This involves getting your hands dirty.

You can see this worked out carefully in this paper by Svante Janson (pp. 1-6). What he calls $u$ and $v$ are, up to scalar factors, what you call $\lambda$ and $\mu$. In his notation, $u+v$ is one of the roots of your polynomial, $uv=-p/3$ (so $u^3v^3=-p^3/27$), and $u^3+v^3$ is the product of the roots, i.e. $-q$. This will give you your explicit quadratic polynomial with roots $u^3$ and $v^3$. Janson takes a slightly different approach, showing that $u^3-v^3=\sqrt{-\Delta/27}$ where the discriminant $\Delta=-4p^3-27q^2$. You can get $u^3$ and $v^3$ directly from those two equations.

To convert the notation, I believe that $v=\lambda/3$ and $u=w^2\mu/3$ if you take $r=\beta_1$,$s=\beta_2$,and $t=\beta_3$.

A good historical reference is Chapter 6 or William Dunham's "Journey through Genius," which covers the discovery of "Cardano's formula," which is what this is... without the Galois theory, because, well he did it about 250 years before Galois. It was one of the first significant results in mathematics that went beyond the legacy of the Greeks.

Answer 2

Viete's theorem provides the answer to your question.

This theorem inverts the idea of solving the quadratic equation (and all higher power polynomial equations) for the roots in terms of the coefficients. Instead, we use the roots of the quadratic equation to find coefficients for an equation that would have these roots. Easy way to see this is to multiply $(x - r_1)(x - r_2)$ so you get a monic quadratic $ax^2+bx+c = 0$ where the coefficients are $a= 1, b= -(r_1+r_2)$ and $c= r_1 r_2$. ($r_1$ is the first root and $r_2$ is the second root of the quadratic equation).

Now imagine $r_1 = \lambda^3$ and $r_2 = \mu^3$. Then your quadratic equation is $$x^2 - (\lambda^3 + \mu^3)x + \lambda^3\mu^3 = 0.$$

Answer 3

I use this formula all the time. If you have found the roots of the resolvent quadratic. Then all you have to do is find the cube roots of x1^3 and x2^3. Your answer should be $(-a+x1^(1/3)+x2^(1/3))/3.$ The other two roots are $(-a+w*x1^(1/3)+w^2*x2^(1/3))/3$ and $(-a+w^2*x1^(1/3)+w*x2^(1/3))/3.$ Where w=(-1+sqrt(-3))/2 and w^2=(-1-sqrt(-3))/2.