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It is easy to convert the general cubic into the form $$y^3+py+q=0$$ via the Tschirnhaus transformation. However, afterwards we can derive with a lot of work Cardano's formula: $$y=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$$

In my book on Galois theory by Stewart, he recalls every nonzero complex number has $3$ cube roots. If one of those cube roots is $\alpha$, then the other two are $\omega\alpha$ and $\omega^2\alpha$ where $\omega=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$. This is fine. However then, Stewart says directly proceeding this: "The expression for $y$ therefore appears to lead to nine solutions, of the form $$\alpha+\beta,\space\space \alpha+\omega\beta,\space\space \alpha+\omega^2\beta$$ $$\omega\alpha+\beta,\space\space \omega\alpha+\omega\beta,\space\space \omega\alpha+\omega^2\beta$$ $$\omega^2\alpha+\beta,\space\space \omega^2\alpha+\omega\beta,\space\space \omega^2\alpha+\omega^2\beta$$ where $\alpha,\beta$ are specific choices of cube roots."

I am confused how he makes the jump from Cardano's formula and a reasoning about every $z\in\mathbb{C}\setminus\{0\}$ having $n$ distinct $n\text{-th}$ roots to this permutation argument.

Stewart further argues that not all of the above expressions are zeros (by the Fundamental Theorem of Algebra) and because we let $\space3\sqrt[3]{u}\sqrt[3]{v}+p=0$ in our derivation of Cardano's formula, we should choose $\alpha,\beta$ so that $3\alpha\beta+p=0,$ then the solutions are $$\alpha+\beta,\space\space \omega\alpha+\omega^2\beta, \space\space\omega^2\alpha+\omega\beta$$

How does Stewart arrive at this permuted list of $9$ solutions and further, how does he narrow them down to the three above?

I hope since these two questions are so very closely related in this section of the book I'll be able to ask them both in the same question here. Thanks!

timon92
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coreyman317
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    Each of cubic root can be chosen in $3$ ways; you have to choose one cubic root for each. that leads to $3^2 = 9$ possible choices overall. If you pick $\alpha$ to be one square root, the other two choices you can make for that square root are $\omega\alpha$ and $\omega^2\alpha$. So that gives you the 9 possibilities listed; the columns leave the second choice fixed and goes through all possible choices of the first cubic root; each row fixes the choice of first cubic root and goes through the choices for three cubic roots. – Arturo Magidin Jan 31 '20 at 19:56
  • As to how he came down to the three choices, he says how: if one choice is $x$ and the other choice is $y$, then you need $xy$ to be a real number (so that $3xy+p=0$). So if we let $\alpha$ and $\beta$ be the real roots, then the only ways in which $x=\omega^i\alpha$ and $y=\omega^j\beta$ have a real product is if $i+j$ is a multiple of $3$, which leads to the three choices he gives. – Arturo Magidin Jan 31 '20 at 20:06
  • I had a problem solving a cubic here and the final three solutions were $$n_0=2\sqrt{\frac{m^2}{3}}\cos\biggl({\biggl(\frac{1}{3}\biggr)\arccos{\biggl(-\frac{3\sqrt{3}D}{2m^4}\biggr)}\biggr)}$$ $$n_1=2\sqrt{\frac{m^2}{3}}\cos\biggl({\biggl(\frac{1}{3}\biggr)\arccos{\biggl(\frac{3\sqrt{3}D}{2m^4}\biggr)}\biggr)}$$ $$n_2=n_1-n_0$$ where $$\lfloor\sqrt[4]{D}\rfloor\le m\le \lceil\sqrt[3]{D}\space \rceil$$ I don't know if any of this applies to you but good luck. – poetasis Jan 31 '20 at 20:55

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We obtain the roots by noting for each root $y$ some $u,\,v\in\Bbb C$ satisfy$$u+v=y,\,uv=-\frac{p}{3},$$from which we can prove$$u^3+v^3=-q,\,u^3v^3=-\frac{p^3}{27}.$$Then $u^3,\,v^3$ are the roots of$$t^2+qt-\frac{p^3}{27}=0,$$and the first root you listed is$$\sqrt[3]{u^3}+\sqrt[3]{v^3}.$$Every root is some $u$ + some $v$, i.e. of the form$$\sqrt[3]{u^3}\omega^j+\sqrt[3]{u^3}\omega^k.$$Without loss of generality we can set $j,\,k\in\{0,\,1,\,2\}$, but there are $3$ roots instead of $9$ (no surprise for a cubic!) because $uv=-\frac{p}{3}$ ensures the choice of $j$ fixes the choice of $k$.

J.G.
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