Showing $\mathbb{Q}(2^{1/3}+2^{1/5})=\mathbb{Q}(2^{1/3},2^{1/5})$ In order to prove $[\mathbb{Q}(2^{1/3}+2^{1/5}):\mathbb{Q}]=15$, I want to show $\mathbb{Q}(2^{1/3}+2^{1/5})=\mathbb{Q}(2^{1/3},2^{1/5})$. Any suggestions?
 A: $F:=\mathbb Q(2^{1/3}+2^{1/5})$ is a subfield of $E:=\mathbb Q(2^{1/3},2^{1/5})=\mathbb Q(2^{1/15})$.
Assume $F\ne E$.
Since $[E:\mathbb Q]=15$, we conclude $[F:\mathbb Q]=3$ or $=5$.
Clearly $2^{1/3}\notin F$ as otherwise also $2^{1/5}=(2^{1/3}+2^{1/5})-2^{1/3}\in F$ an dthus $F=E$.
Similarly, $2^{1/5}\notin F$.
Assume $[F:\mathbb Q]=3$. Then $1<[F(2^{1/3}):F]\le 3$  and $[F(2^{1/3}):F]\cdot [E:F(2^{1/3})]=5$, contradiction.
Assume $[F:\mathbb Q]=5$. Then from $1<[F(2^{1/5}):F]\le 5$  and $[F(2^{1/5}):F]\cdot [E:F(2^{1/5})]=3$, we obtain $[F(2^{1/15}):F]=3$.
A: Let $K=\mathbb{Q}(\zeta_{15})$, $L=K(2^{1/3}+2^{1/5})$, and $M=K(2^{1/5},2^{1/3})=K(2^{1/15})$.  One can easily check $M/K$ is cyclic Galois of degree 15.  Since $2^{1/5}=2^{1/3}+2^{1/5}-2^{1/3}$, $M=L(2^{1/3})$, so $M/L$ has degree at most 3.  Because $[M:L]$ must divide 15, $M/L$ has degree 1 or 3, and $L/K$ has degree 5 or 15.
Suppose $L/K$ has degree 5.  Since $M/K$ is cyclic, there is only one degree 5 subextension, so $L=K(2^{1/5})$.  Then $2^{1/3}=2^{1/3}+2^{1/5}-2^{1/5}$ lies in $L$, so $L=M$ which contradicts the fact that $L/K$ has degree $5$.  Thus $L/K$ has degree 15.  Hence $2^{1/3}+2^{1/5}$ doesn't satisfy a polynomial of degree less than 15 over $K$.  
Since $\mathbb{Q}\subseteq K$, it follows that the minimal polynomial over $\mathbb{Q}$ has degree at least $15$.  Hence $[\mathbb{Q}(2^{1/3}+2^{1/5}):\mathbb{Q}]\geq 15$.  But the only subextension of $\mathbb{Q}(2^{1/3},2^{1/5})/\mathbb{Q}$ with degree at least 15 is $\mathbb{Q}(2^{1/3},2^{1/5})$, so $\mathbb{Q}(2^{1/3}+2^{1/5})=\mathbb{Q}.(2^{1/3},2^{1/5})$.
