Two question about ${\mathbb Q}(\alpha)$ for $\alpha$ := $\sqrt[n]{7}$ + $\sqrt[n+3]{7}$ Let $\alpha$ = $\sqrt[n]{7}$ + $\sqrt[n+3]{7}$ with $n$ not divisible by $3$.


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*Prove that $[{\mathbb Q}(\alpha) : {\mathbb Q}] = n(n + 3)$.

*Conclude that $\alpha$ is constructible if and only if $n = 1$, and for $n = 1$ show in
words or in illustrations how to construct $\alpha$.
I have an assignment due tomorrow so any help would be superb. zero idea to be perfectly honest
 A: First, recall that $x^m-p$ is always irreducible over $\mathbb{Q}$ for $p$ prime (via Eisenstein's criterion).
Consider the field $\mathbb{Q}[\sqrt[n]{7},\sqrt[n+3]{7}]$. It contains $\mathbb{Q}[\sqrt[n]{7}]$ and $\mathbb{Q}[\sqrt[n+3]{7}]$ so its dimension is divisible by $n$ and $n+3$. They are coprime (since $3\nmid n$) so the dimension is at least $n(n+3)$ and clearly it can't be larger so it is exactly $n(n+3)$, thus giving you an upper bound for$[\mathbb{Q}[\sqrt[n]{7}+\sqrt[n+3]{7}]:\mathbb{Q}]$.
For the following I will assume that all the roots considered in your question are the positive real roots (though I'm not sure if this condition is necessary).
Suppose that $f(x)=x^n+7=h(x)g(x)$ is reducible over the field $L=\mathbb{Q}[\sqrt[n]{7}+\sqrt[n+3]{7}]$. Since its roots are $\zeta_n^i \sqrt[n]{7}$, we get that $h(0)=\pm \zeta_n^j 7^{\deg(h)/n}\in L$ for some $j$. Since the field is real, we have that $\zeta_h^i = \pm 1$ so that $7^{\deg(h)/n}\in L$. Writing $\deg(h)=d\cdot a$ with $d=gcd(\deg(h),n)$ and $gcd(a,n)=1$, we can find some $b$ such that $ab\equiv_n 1$. We conclude that $(7^{\deg(h)/n})^b=7^{ab\cdot d / n}=7^k \cdot 7^{1/(n/d)}\in L$ for some integer $k$ and therefore $\sqrt[n/d]{7}\in L$. In other words, we have found a polynomial of degree $d\leq \deg(h)$ over $L$ that has the root $\sqrt[n]{7}$. It follows that the minimal polynomial must have this form, and its degree divides $n$. We now have
$$n(n+3)=[\mathbb{Q}[\sqrt[n]{7},\sqrt[n+3]{7}]:\mathbb{Q}]=[L[\sqrt[n]{7}]:\mathbb{Q}]=[L[\sqrt[n]{7}]:L] \cdot [L:\mathbb{Q}]=d\cdot [L:\mathbb{Q}]$$ 
and we conclude that $n+3 \mid [L:\mathbb{Q}]$. Similarly we have that $n\mid [L:\mathbb{Q}]$ so that $(n+3)n \mid [L:\mathbb{Q}]$ and there must be equality. If you find a better argument for this step, I will be happy to hear about it.
For your second question, I will just hint that constructible numbers always come from iterations of quadratic extensions.
