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?

Just a remark/simplification: if the polynomial $x^3 - 3$ is not irreducible over $\mathbb{Q}(\sqrt[3]{2})$, then at least one of the factors must be linear (this is a special nice feature of degree $\leq 3$ of course). So, it is enough to show that this polynomial has no roots in $\mathbb{Q}(\sqrt[3]{2})$, you don't need to check the quadratic case separately. The roots of that polynomial over $\overline{\mathbb{Q}}$ are of course $\omega^i\sqrt[3]{3}$, where $\omega$ is a primitive cube root of unity and $0\leq i\leq 2$. So, you just apply the first step of your argument to all these roots at the same time and save yourself the second step.

Edit: To finish the first step, suppose for a contradiction that you have $$ \begin{align} a^3 + 2b^3 + 4c^3 + 12abc & = & 3,\;\;\;\;\;\;\;\;\;\;\;(1) \\ a^2b+2b^2c + 2c^2a & = & 0,\;\;\;\;\;\;\;\;\;\;\;(2)\\ a^2c+b^2a+2c^2b & = & 0.\;\;\;\;\;\;\;\;\;\;\;(3) \end{align} $$ First, multiply through by the lcm of the denominators of $a,b,c$ to make them integers, so the equations become $$ \begin{align} a^3 + 2b^3 + 4c^3 + 12abc & = & 3d^3,\;\;\;\;\;\;\;\;\;\;\;(1) \\ a^2b+2b^2c + 2c^2a & = & 0,\;\;\;\;\;\;\;\;\;\;\;(2)\\ a^2c+b^2a+2c^2b & = & 0.\;\;\;\;\;\;\;\;\;\;\;(3) \end{align} $$ We may assume without loss of generality that $a,b,c$ are coprime, otherwise divide through by the highest common factor.

Step 1: First, I claim that $a$ (and therefore also $d$, but that doesn't matter) is odd. Indeed, let's look at equation (1): $$ 2|a\Rightarrow 8|RHS\Rightarrow \text{ and }LHS\equiv 2b^3\pmod 4\Rightarrow 2|b\Rightarrow LHS\equiv 4c^3\pmod 8\Rightarrow 2|c. $$ This contradicts coprimality of $a,b,c$, so $a$ is odd.

Step 2: With this information, we will consider (2) and (3) modulo higher and higher powers of 2: (2) modulo 2 forces $b$ to be even. This in turn means that (3) forces $c$ to be even.

Now, we will alternate between the two equations. If $c$ is even then (2) $\Rightarrow4|b$. Thus, (3) $\Rightarrow4|c$. But then (2) $\Rightarrow 8|b$. And so on. This will never stop. Since $b$ and $c$ cannot be divisible by arbitrarily large powers of 2, we obtain a contradiction.

This is a variant of Fermat's infinite descent. Note that I could have phrase Step 2 in a finite way: write $b=2^nu$, $c=2^mv$ with $u$ and $v$ odd, obtain inequalities between $n$ and $m$ that contradict each other. Equally well, I could have phrased Step 1 in the language of infinite descent, without assuming coprimality at the outset.

However, the cleanest formulation would have been obtained by using $p$-adic valuations. Recall that given a prime $p$ and a rational number $x$, if we write $x=p^n\frac{r}{s}$, where $n\in \mathbb{Z}$ and $p\nmid r,s$, then the $p$-adic valuation of $x$ is defined as $\text{val}_p(x) = n$. The fundamental property of the valuation is that $$ \text{val}_p\left(\sum_i x_i\right)\geq \text{min}(\text{val}_p(x_i))_i;\;\;\;\;\;\;\;(4)$$ moreover, if there exists $x_0$ such that $\text{val}_p(x_0)<\text{val}_p(x_i)$ for all $i\neq 0$, then we have equality in (4). I propose it as an exercise for the reader to rephrase both steps above in the language of valuations, using these two properties. Note, that you then don't care whether $a,b,c$ are integers or rationals.

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.