Minimal polynomial of the element $1+\sqrt[3]{2}+\sqrt[3]{4}$ Over $Q$

Question

I guess $Q(1+\sqrt[3]{2}+\sqrt[3]{4}) = Q(\sqrt[3]{2})$ hence degree of minimal polynomial should be 3. But I am not able to see this.

Answer 1

The degree is at most $3$, so the elements $1$, $\alpha$, $\alpha^2$ and $\alpha^3$ are linearly dependent over $\Bbb{Q}$. Compute these powers and find the linear dependence by basic linear algebra.

Alternatively note that $$(1+\sqrt[3]{2}+\sqrt[3]{4})(\sqrt[3]{2}-1)=1,$$ and the minimal polynomial of $\sqrt[3]{2}-1$ is clearly $$(x+1)^3-2=x^3+3x^2+3x-1,$$ so the minimal polynomial of your element is $$-x^3((x^{-1}+1)^3-2)=x^3-3x^2-3x-1.$$

Answer 2

$1+\sqrt[3]2+\sqrt[3]4$ is an eigenvalue of $A=\pmatrix{1&1&1\\2&1&1\\2&2&1}$. In fact $$\pmatrix{1&1&1\\2&1&1\\2&2&1}\pmatrix{1\\\sqrt[3]2\\\sqrt[3]4} =(1+\sqrt[3]2+\sqrt[3]4) \pmatrix{1\\\sqrt[3]2\\\sqrt[3]4}.$$ So just compute the characteristic polynomial of $A$ to get the minimum polynomial of $1+\sqrt[3]2+\sqrt[3]4$.