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There are only answers without any reasons in my textbook's questions.

PLEASE HELP ME :(

Find the root of these polynomials by using hints.

Let the $w$ is complex root of $x^2+x+1$

1.First question

$f(x)=x^3 + 6x^2 -2 $ which means $ irr(\alpha,Q)$ (hint : one of the real root of $f(x)$ is $\alpha = \sqrt[3]{4}-\sqrt[3]{2}$ )

Solution in my textbook said that that all the roots are

$\alpha$

$\beta$ =$\sqrt[3]{4}w^2-\sqrt[3]{2}w$,

$\gamma$ = $\sqrt[3]{4}w-\sqrt[3]{2}w^2$'

I'm tried to find the other roots $\beta$ and $\gamma$ But failed. :(

Is there any Either formula or principle finding the $\beta$ and $\gamma$? Please give me some ideas.

2.Second question

$g(x)=x^9 -3x^6+165x^3-1 $ which means $ irr(\alpha,Q)$ (hint : one of the real root of $g(x)$ is $\alpha = \sqrt[3]{3}-\sqrt[3]{2}$ )

Solution in my textbook said that that all the roots form is

$ \sqrt[3]{3}w^n-\sqrt[3]{2}w^m$ [ $m,n=0,1,2$]

All I know the number of roots is 9 thinking the splitting field of $g(x)$ over $Q$ And It might having the similar principle with the case of $f(x)$

BUT I couldn't find exact all the complex roots though $\alpha$ was given.

Why is all the roots of $g(x)$ have a form that $ \sqrt[3]{3}w^n-\sqrt[3]{2}w^m$?

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    $\begingroup$ Can you use the Cardano's formulas ? $\endgroup$ – Yves Daoust May 10 at 9:51
  • $\begingroup$ Use $\omega$ \omega rather than $w$. $\endgroup$ – Yves Daoust May 10 at 9:52
  • $\begingroup$ I used cardano's formula but I could solve only first question. It is not useful when thinking the second case. $\endgroup$ – se-hyuck yang May 10 at 10:19
  • $\begingroup$ Are you so sure ? $\endgroup$ – Yves Daoust May 10 at 10:46
  • $\begingroup$ When applying formula, the first question is solved easily. But the second case when substituting x as t^3 it was really hard. I want to know the reason why the second case have a form like above rather than the formula. $\endgroup$ – se-hyuck yang May 10 at 10:52
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Let $\theta = \sqrt[3]{2}$. Then the conjugates of $\theta$ are $\theta\omega$ and $\theta\omega^2$, where $\omega$ is a primitive cubic root of unity.

Now $\alpha = \theta^2-\theta = h(\theta)$ and so its conjugates are $h(\theta\omega)$ and $h(\theta\omega^2)$. These are the other roots of $f$.

A similar answer works for the second question.

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  • $\begingroup$ Simple method! But Could you more explain the reason $h(\theta)$, $h(\theta \omega)$ and $h(\theta \omega^2)$ are conjugate each other? $\endgroup$ – se-hyuck yang May 10 at 11:44
  • $\begingroup$ Plus I thought since the each roots mapped by $h$, so the $x^3-2$ mapped $h$ which means $h(x^3-2)$. Then $h(x^3-2)$ is not $f$. So the root is not equal. It might the false thinking, but I can't find which point I was wrong. $\endgroup$ – se-hyuck yang May 10 at 11:56
  • $\begingroup$ @se-hyuckyang, they are conjugate because $h$ is a polynomial with rational coefficients and so commutes with every automorphism that fixes $\mathbb Q$. $\endgroup$ – lhf May 10 at 12:13
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Divide $$(x^3+6x^2-2):(x-(\sqrt[3]{4}-\sqrt[3]{2}))=$$ and you will get a quadratic one.

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  • 2
    $\begingroup$ You mean divide by $x-...$? $\endgroup$ – J. W. Tanner May 10 at 12:02
  • $\begingroup$ and $6x^\color{red}2$? $\endgroup$ – J. W. Tanner May 10 at 18:10

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