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?
 A: 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.
A: 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.
A: I'm appalled at the amount of work it seems to take to get what should be an "obvious" result...that the cube root of $3$ does not magically appear when you extend the rationals by the cube root of $2$. I'm going to try to look for consequences of such an unlikely circumstance and see if we can get a contradiction.
First, let's let $\alpha=\sqrt[3]2$  and  $\beta = \sqrt[3]3,$ where we will understand the cube roots to refer to the real values. The undesirable outcome is to suppose that
$$\beta = a + b\alpha + c\alpha^2$$
But let’s remember that there are three cube roots for every number; and since they are algebraically indistinguishable, what is true for one must be true for all. The obvious interpretation of this equation is that it applies to the real cube roots; but what if we try applying it to the complex values? Let’s try the substitution of $\omega\alpha$ for $\alpha$; this naturally forces the substitution of $\omega^2\beta$ for $\beta$. What does the equation look like now?
$$(\omega^2) \beta = a + b\omega\alpha + c (\omega^2)\alpha^2$$
This equation is oddly problematic. Assuming as we may that $\alpha$ and $\beta$ are real, we can interpret this quantity as a vector in the complex plane. From the LHS, it points in the direction of $\omega^2$; but on the RHS, it is made up of three vectors, one of which is already pointing in the right direction. In order for the whole vector to be pointing the right way, the remaining components must also be pointing in the direction of $\omega^2$:
$$a + b\omega\alpha =k(\omega^2)$$
It is clear from geometrical inspection that this can only happen if $a=b\alpha$; and since $a$ and $b$ are rational, and $\alpha$ is irrational, this is impossible. Therefore, going back to the original assumption, we must find that $\beta =\sqrt[3]3$ is not present in the extension field $\Bbb Q(\alpha=\sqrt[3]2)$.
I wonder if this argument is correct, or "close enough" to correct that it can be fixed up? I am a little uncomfortable with the general flavor of an algebraic argument which resorts to geometric interpretations; it's not the way I expect these things to go.   
A: As was observed in the OP, it suffices to show that the polynomial $f(x) = x^9 - 15x^6 - 87x^3 -125$ is irreducible (over $\mathbb{Q}$). The Eisenstein Criterion with p = 5 doesn't work with this polynomial. However (with help from Wolfram Alpha), we find that
$f(x - 1) = x^9-9 x^8+36 x^7-99 x^6+216 x^5-438 x^4+471 x^3-261 x^2+99 x-141$
and the Eisenstein Criterion with p = 3 shows that f(x - 1), and hence f(x), is irreducible over $\mathbb{Q}$.
I'm assuming that the OP knows that $\mathbb{Q}(\sqrt[3]2, \sqrt[3]3) = \mathbb{Q}(\sqrt[3]2 + \sqrt[3]3)$.
