Polynomial in $\mathbb Z[X]$ with a peculiar root I saw this problem on a website a while ago and I 'm still stuck.
Let $\alpha = \large\sqrt[7] \frac{3}{5}+\sqrt[7] \frac{5}{3}$.
Find and prove uniqueness of a polynomial $P \in \mathbb Z[X]$, with degree $7$ and leading coefficient $-15$ such that $P(\alpha)=0$
It definitely has something to do with properties of $r+\frac{1}{r}$.
Any hint is welcome.
 A: We may use the identity
$$\eqalign{
(X+Y)^7-X^7-Y^7&=7XY(X+Y)(X^2+XY+Y^2)^2\cr
&=7XY(X+Y)((X+Y)^2-XY)^2
}
$$
So choosing $X=\root{7}\of{3/5}$ and $Y=1/X$ so that $\alpha=X+Y$ we get
$$
\alpha^7-\frac{3}{5}-\frac{5}{3}=7\alpha(\alpha^2-1)^2
$$
and this reduces to
$$
34+105\alpha-210\alpha^3+105\alpha^5 -15\alpha^7=0.
$$
Now, let us prove that 
$$
P(X)=34+105X-210X^3+105X^5 -15 X^7
$$
is Irreducible in $\Bbb{Z}[X]$ (or equivalently in $\Bbb{Q}[X]$). Indeed, If
$$Q(X)=X^7P(1/X)=34X^7+105X^6-210X^4+105X^2 -15$$
Then clearly $5$ divides all the coefficients of $Q$ except the leading one, and $25$ does not divide the constant term. This proves that $Q$ is irreducible in $\Bbb{Z}[X]$ according to Eisenstein Irreducibility criterion. So the same holds for $P$.
Now the fact that $P$ is irreducible $\Bbb{Z}[X]$, proves that it is the minimal polynomial of $\alpha$ (because otherwise the minimal polynomial would be a factor of $P(X)$), and this minimality implies  the uniqueness statement of your question. $\qquad\square$
