# Find all roots of these two polynomials

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

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 $$\sqrt[3]{3}w^n-\sqrt[3]{2}w^m$$?

• Can you use the Cardano's formulas ? – Yves Daoust May 10 at 9:51
• Use $\omega$ \omega rather than $w$. – Yves Daoust May 10 at 9:52
• I used cardano's formula but I could solve only first question. It is not useful when thinking the second case. – se-hyuck yang May 10 at 10:19
• Are you so sure ? – Yves Daoust May 10 at 10:46
• 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. – se-hyuck yang May 10 at 10:52

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.

• Simple method! But Could you more explain the reason $h(\theta)$, $h(\theta \omega)$ and $h(\theta \omega^2)$ are conjugate each other? – se-hyuck yang May 10 at 11:44
• 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. – se-hyuck yang May 10 at 11:56
• @se-hyuckyang, they are conjugate because $h$ is a polynomial with rational coefficients and so commutes with every automorphism that fixes $\mathbb Q$. – lhf May 10 at 12:13

Divide $$(x^3+6x^2-2):(x-(\sqrt[3]{4}-\sqrt[3]{2}))=$$ and you will get a quadratic one.

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