<s>[INST] How to show that $[\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{3}):\mathbb{Q}]=9$? 
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?
 [/INST] Since this question has been bugging me for a few years, I decided to go ahead and chug out the calculations necessary to show "$q$ and $r$ can't both be zero," as Gerry Myerson said.
Since from your cubing, we get
$$p=a^3+2b^3+4c^3+12abc$$
$$q=3a^2b+6b^2c+6c^2a$$
$$r=3b^2a+6c^2b+3a^2c$$
and in Gerry Myerson's case we would have the second and third of these equations equal to zero and thus the first would have to be 3, we get
$$a^3+2b^3+4c^3+12abc=3$$
$$3a^2b+6b^2c+6c^2a=0$$
$$3b^2a+6c^2b+3a^2c=0$$
and the goal is to show that this can't happen with rational $a,b,c.$
Dividing the second and third equations by 3 and completing the square, we find that
$$(ba+c^2)^2=c^4-2b^3c$$
$$(ab+c^2)^2=c^4-ca^3$$
so that $a^3=2b^3.$ Then solving the first of the three equations for $abc,$ we get
$$abc=\frac{3-4b^3-4c^3}{12}.$$
Multiplying the second of the three equations through by $b,$ and substituting for $abc,$ we find that
$$6a^2b^2+8b^3c+3c-4c^4=0.$$
This is ready to be multiplied through by $c^2,$ and then substituting for $abc$ once again, we get the equation of the sixth degree in $b$ and in $c$
$$16b^6+(224c^3-24)b^3+12c^6+48c^3+9=0.$$
We can let $B=b^3, C=c^3,$ and then this becomes a quadratic, where you can view one of the variables (say $B$) as what we're solving for and the other as a parameter. Remembering that the quadratic would need to have a rational square for a discriminant in order to have a rational root, we find that
$$(224C-24)^2-4\cdot16\cdot(12C^2+48C+9)=r^2$$
for some rational number $r.$ Now we can view this as a quadratic in $C$ and apply the same discriminant trick, and find that
$$4\cdot240^2+4\cdot193r^2=s^2$$
for a rational number $s.$ Letting $r=m/n,$ where $m,n$ are integers,
$$240^2 n^2 + 193 m^2 = (ns/2)^2.$$
Mod 5, this says $3m^2$ is a square, which is impossible.
I hope someone who knows Galois theory well can provide a more easily generalizable answer.
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