Roots of Resolvent of Cubic

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $P$ be the cubic equation:

$a x^3 + b x^2 + c x + d = 0$ with $a \ne 0$

Let $\map r P$ be the resolvent equation of $P$, given by:

$u^6 - 2 R u^3 - Q^3$

Let the roots of $P$ be $\alpha_1, \alpha_2, \alpha_3$.


Then the roots of $\map r P$ can be expressed as:

\(\ds v\) \(=\) \(\ds \frac 1 3 \paren {\alpha_1 + \omega \alpha_2 + \omega^2 \alpha_3}\)
\(\ds \omega v\) \(=\) \(\ds \frac 1 3 \paren {\alpha_3 + \omega \alpha_1 + \omega^2 \alpha_2}\)
\(\ds \omega^2 v\) \(=\) \(\ds \frac 1 3 \paren {\alpha_2 + \omega \alpha_3 + \omega^2 \alpha_1}\)
\(\ds u\) \(=\) \(\ds \frac 1 3 \paren {\alpha_1 + \omega \alpha_3 + \omega^2 \alpha_2}\)
\(\ds \omega u\) \(=\) \(\ds \frac 1 3 \paren {\alpha_2 + \omega \alpha_1 + \omega^2 \alpha_3}\)
\(\ds \omega^2 u\) \(=\) \(\ds \frac 1 3 \paren {\alpha_3 + \omega \alpha_2 + \omega^2 \alpha_1}\)

where $\omega = -\dfrac {-1 + \sqrt {-3} } 2$ is one of the primitive (complex) cube roots of $1$.


Proof


This theorem requires a proof.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{ProofWanted}} from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Roots_of_Resolvent_of_Cubic&oldid=373173"