Fraleigh, Sec31, Ex9. Show that $[\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{3}):\mathbb{Q}]=9$.
Here is my trial: It is obvious that $\sqrt[3]2$ is algebraic of degree 3 over $\mathbb{Q}$, since $x^3-2$ is irreducible over $\mathbb{Q}$ by Eisenstein crieterion with $p=2$. Then we need to show that $\sqrt[3]3$ is algebraic of degree 3 over $\mathbb{Q}(\sqrt[3]2)$. Since $\sqrt[3]3$ is a zero of $x^3-3$, its degree is at most 3.
To show that $\sqrt[3]3$ is not of degree 1, i.e. $\sqrt[3]3 \not\in \mathbb{Q}(\sqrt[3]2)$, suppose that $\sqrt[3]3=a+b\sqrt[3]2+c\sqrt[3]4$, where $a, b, c \in \mathbb{Q}$. (The usual degree argument is not available since $\deg(\sqrt[3]3,\mathbb{Q})=3$ divides $\deg(\sqrt[3]2,\mathbb{Q})=3$.) Cubing both sides, $3=p+q\sqrt[3]2+r\sqrt[3]4$ with some $p, q, r$ in $\mathbb{Q}$, so $\sqrt[3]2$ is a zero of $rx^2+qx+p-3$, which is a contradiction to $\deg(\sqrt[3]2,\mathbb{Q})=3$.
Now to prove $\sqrt[3]3$ is not of degree 2, suppose that $\sqrt[3]3$ is a zero of quadratic polynomial. This means that $\sqrt[3]9=p\sqrt[3]3+q$ for some $p,q \in \mathbb{Q}(\sqrt[3]2)$. Cubing both sides, $9=3p^3+q^3+3p\sqrt[3]3q(\sqrt[3]3p+q)=3p^3+q^3+9pq:=a+b\sqrt[3]2+c\sqrt[3]4$ for some $a,b,c\in\mathbb{Q}$, which leads to the same contradiction.
Actually I didn't know how to solve it but while writing out this question, it seems that I solved the problem. But is the above solution right? And is there any other way to solve it? At first, I tried to show that $x=\sqrt[3]{2}+\sqrt[3]{3}$ is algebraic of degree 9 over $\mathbb{Q}$. Cubing yields that $x^3=5+3 \sqrt[3]{6}x$, so $(x^3-5)^3=162x^3$, so $x^9-15x^6-87x^3-125=0$ has $\sqrt[3]{2}+\sqrt[3]{3}$ as a zero. But I couldn't show that it is irreducible (Eisenstein criterion with $p=5$ fails.)
Edit: As Alex pointed out, it is sufficient to show that $x^3-3$ has no roots in $\mathbb{Q}(\sqrt[3]{2})$. And as Gerry pointed out, this process some more work than the above(check the nonzero condition). Suppose $(a+b\sqrt[3]2+c\sqrt[3]4)^3=3$. I did the heavy computation, $a^3+2b^3+4c^3+12abc+3\sqrt[3]2(a^2b+2b^2c+2c^2a)+3\sqrt[3]4(a^2c+b^2a+2c^2b)=3$, and stuck on here. How can I proceed here?