Determine the degree of $1+\sqrt[3]{2}+\sqrt[3]{4}$ over $\mathbb{Q} ?$

Question

I have to find the degree of $1+\sqrt[3]{2}+\sqrt[3]{4}$ over $\mathbb{Q}$. This is what I found already:

$\mathbb{Q}(1+\sqrt[3]{2}+\sqrt[3]{4})=\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{4})$

$[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$ and $[\mathbb{Q}(\sqrt[3]{4}):\mathbb{Q}]=3$.

Hence $k=[\mathbb{Q}(1+\sqrt[3]{2}+\sqrt[3]{4}):\mathbb{Q}]\leq9$. Also $3|k$. So $k$ can be $3,6$ or $9$. How can I find the exact value of $k$? Can anyone please help? Have I missed some proposition that can be directly applied?

Answer 1

Let $x = 1+\sqrt[3]{2} + \sqrt[3]{4} \to x = \dfrac{1}{\sqrt[3]{2}-1} \to \sqrt[3]{2} = \dfrac{1+x}{x} \to (1+x)^3 = 2x^3 \to x^3 - 3x^2 - 3x - 1 = 0 \to \text{deg} = 3$