# Prove $\sqrt[3]{3} \notin \mathbb{Q}(\sqrt[4]{3})$

The question asks us to prove $\sqrt[3]{3} \notin \mathbb{Q}(\sqrt[4]{3})$

I realise this is similar to another question - Show that $\sqrt{2}\notin \mathbb{Q}(\sqrt[4]{3})$ - and in this question the answer recommends using the fact that $\mathbb{Q}\subset\mathbb{Q}(\sqrt{3})\subset\mathbb{Q}(\sqrt[4]{3})$, which in turn means that if we show that $\sqrt[3]{3} \notin \mathbb{Q}(\sqrt{3})$, we can then go on to show that it is not in $\mathbb{Q}(\sqrt[4]{3})$.

Following these steps, I said that $\sqrt[3]{3} = a+b\sqrt{3}$ and then cubing both sides and subtracting 3, got that $(a^3+9ab^2-3)+3\sqrt{3}(a^2b+b^3)=0$ and this is eventually where I got stuck.

You can prove this using the tower law. Suppose towards a contradiction that the claim is true. Then we have the tower $$\mathbb{Q}\subset\mathbb{Q}(\sqrt[3]{3})\subset\mathbb{Q}(\sqrt[4]{3}).$$ Note $\sqrt[3]{3}$ has degree $3$ over $\mathbb{Q}$ while $\sqrt[4]{3}$ has degree $4$ or $2$ over $\mathbb{Q}$ (doesn't matter which). But $3$ does not divide $4$ and does not divide $2$. Contradiction.
From $(a^3+9ab^2-3)+3\sqrt{3}(a^2b+b^3)=0$ and the fact that $a, b$ are rational, we get $a^3 + 9ab^2 = 3$ and $a^2b + b^3 = 0$. Since $b \neq 0$, we get $a^2 + b^2 = 0$ and hence $a=b=0$, a contradiction.
Since $a,b$ are rational, $0$ is rational and $\sqrt3$ is irrational, we derive from the second term of your final equation that $a^2b+b^3=0$. Factoring out $b$, this means either $b=0$ or $a^2=-b^2$, which again implies $b=0$. Carrying $b=0$ to the first term of the equation, we get $a^3-3=0$, which is unsolvable, since we want $a$ to be rational.
• So, now having proved that $\sqrt[3]{3} \notin \mathbb{Q}(\sqrt{3})$, the final step would then be to check that these is no number of the form $a+b\sqrt[4]{3}$ with $a,b \in \mathbb{Q}(\sqrt{3})$ that cubes to 3. Would this be correct? – user402253 May 18 '17 at 22:39