# Solving Cubic Equations with Lagrange Resolvent? Here's what I understand: x^3+px-q=(x-r)(x-s)(x-t).

Solving Cubic Equations with Lagrange Resolvent?
I'm having difficulties understanding my textbook's decription of solving cubic equations using Lagrange Resolvents and symmetric polynomials.
Here's what I understand:
${x}^{3}+px-q=\left(x-r\right)\left(x-s\right)\left(x-t\right)$
We can also write:
$\lambda =r+ws+{w}^{2}t$
$\mu =wr+s+{w}^{2}t$
where $1,w,{w}^{2}$ are the cubic roots of 1. I then understand that ${\lambda }^{2}+{\mu }^{3}$ 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 $q\left(x\right)={x}^{2}-\left({\lambda }^{3}+{\mu }^{3}\right)x+{\lambda }^{3}{\mu }^{3}$, 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"?
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Sauppypefpg
Step 1
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 $\left(x-{r}_{1}\right)\left(x-{r}_{2}\right)$ so you get a monic quadratic $a{x}^{2}+bx+c=0$ where the coefficients are $a=1,b=-\left({r}_{1}+{r}_{2}\right)$ and $c={r}_{1}{r}_{2}$. (${r}_{1}$ is the first root and ${r}_{2}$ is the second root of the quadratic equation).
Step 2
Now imagine ${r}_{1}={\lambda }^{3}$ and ${r}_{2}={\mu }^{3}$. Then your quadratic equation is
${x}^{2}-\left({\lambda }^{3}+{\mu }^{3}\right)x+{\lambda }^{3}{\mu }^{3}=0.$
###### Did you like this example?
Krish Logan
Step 1
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 $x{1}^{3}$ and $x{2}^{3}$.
Step 2
Your answer should be $\left(-a+x{1}^{\left(}1/3\right)+x{2}^{\left(}1/3\right)\right)/3.$. The other two roots are $\left(-a+w\ast x{1}^{\left(}1/3\right)+{w}^{2}\ast x{2}^{\left(}1/3\right)\right)/3$ and $\left(-a+{w}^{2}\ast x{1}^{\left(}1/3\right)+w\ast x{2}^{\left(}1/3\right)\right)/3.$. Where .