How to formally prove that the degree of $\Bbb Q（\sqrt[5]{6},\sqrt[3]{7}）$ is $15$？ I tried to consider the tower of extension $\Bbb Q\subset \Bbb Q（\sqrt[3]{7}）\subset\Bbb Q（\sqrt[5]{6},\sqrt[3]{7}）$.
The minimal polynomial of $\Bbb Q（\sqrt[3]{7}）$ over $\Bbb Q$ is $x^3-7$ by Eisenstein. But although it is easy to see that it has no root in $\Bbb Q（\sqrt[3]{7}）$, how can I formally conclude that $x^5-6$ is irreducible over $\Bbb Q（\sqrt[3]{7}）$ and thus we can see the basis of $\Bbb Q（\sqrt[3]{7}, \sqrt[5]{6}）$？
Thanks！
 A: Hint: Consider both extensions $\mathbb{Q}\subset\mathbb{Q}(\sqrt[3]{7})\subset\mathbb{Q}(\sqrt[3]{7},\sqrt[5]{6})$ and  $\mathbb{Q}\subset\mathbb{Q}(\sqrt[5]{6})\subset\mathbb{Q}(\sqrt[3]{7},\sqrt[5]{6})$. Which conditions (multiple of...) must the degree of $\mathbb{Q}\subset\mathbb{Q}(\sqrt[3]{7},\sqrt[5]{6})$ satisfy (using transitivity of degree)? 
(Edit: in other words, draw the two towers next to each other)
A: Since $\mathbb{Q}(\sqrt[5]6,\sqrt[3]7)$ is a finitely generated algebraic extension of $\mathbb{Q}$, it is a finite extension of $\mathbb{Q}$.
Hence $[\mathbb{Q}(\sqrt[5]6,\sqrt[3]7)]=[\mathbb{Q}(\sqrt[5]6,\sqrt[3]7):\mathbb{Q}(\sqrt[5]6)][\mathbb{Q}(\sqrt[5]6):\mathbb{Q}]$.
And also, $[\mathbb{Q}(\sqrt[5]6,\sqrt[3]7)]=[\mathbb{Q}(\sqrt[5]6,\sqrt[3]7):\mathbb{Q}(\sqrt[3]7)][\mathbb{Q}(\sqrt[3]7):\mathbb{Q}]$.
You are correct in your reasoning for $[\mathbb{Q}(\sqrt[3]7):\mathbb{Q}]=3$ and $[\mathbb{Q}(\sqrt[5]6):\mathbb{Q}]=5$.
Now consider $[\mathbb{Q}(\sqrt[5]6,\sqrt[3]7):\mathbb{Q}(\sqrt[5]6)]$.
You will find that the minimum polynomial of $\sqrt[3]7$ over $\mathbb{Q}$, $m(x)$ say, divides $x^3-7$ in $\mathbb{Q}(\sqrt[5]6)$.
Let $deg(m(x))=k$. Then $1\leq k \leq 3$.
Then from the second line in this answer, we then have:
$[\mathbb{Q}(\sqrt[5]6,\sqrt[3]7):\mathbb{Q}]=5k$.
Now from the third line we find, $3|5k$.
But by Euclid's Lemma, $gcd(5,3)=1$. So $3|k$.
But $1\leq k \leq 3$, hence $k=3$.
Thus $[\mathbb{Q}(\sqrt[5]6,\sqrt[3]7):\mathbb{Q}]=3.5=15$.
