$3 X^3 + 4 Y^3 + 5 Z^3$ has roots in all $\mathbb{Q}_p$ and $\mathbb{R}$ but not in $\mathbb{Q}$ This is an exercise in my textbook in a chapter about the Hasse-Minkowski-theorem:

Show that the polynomial $3 X^3 + 4 Y^3 + 5 Z^3$ has a non-trivial root in $\mathbb{R}$ and all $\mathbb{Q}_p$. Show that it has only the trivial root in $\mathbb{Q}$.

I don't know, at the first sight, this seems like a pretty hard exercise, especially the second part. Is it doable? Do you have any tips how to start?
Or should I simply skip it? Because I don't see how this exercise helps me to understand/apply the Hasse-Minkowski-theorem.

This is a classical counterexample to Hasse-Minkowski, due to Selmer. The solution of the second part is indeed difficult. For a detailed proof see, for example, Chapter $7$ of the thesis of Arnélie Schinck. 
However, the first part is quite easy, and goes as follows: we can explicitly list a solution for each $\mathbb{Q}_p$, including for $\mathbb{Q}_{\infty}=\mathbb{R}$:
$$
(x,y,z)=(-1,(3/4)^{1 /3},0),(0,(5/4)^{1/3},-1),(5,-2(15/4)^{1/3},-3),(-1,0,(3/5)^{1/3}).
$$
All solutions exist in $\mathbb{R}$, and at least one exists in a given $\mathbb{Q}_p$.
