Prove that $\sqrt[5]a+\sqrt[5]b+\sqrt[5]c=0$ Let $a$, $b$ and $c$ be roots of the equation
$$x^3+15x^2-198x+1=0.$$
Prove that:
$$\sqrt[5]a+\sqrt[5]b+\sqrt[5]c=0$$
I have a solution for this problem, but I want to see another solutions.
My solution is the following.
We'll take the equation $x^3-3x+1=0$, 
which has three roots $a=2\cos40^{\circ}$, $b=2\cos80^{\circ}$ and $c=-2\cos20^{\circ}$.
Now, easy to show that 
$a^5+b^5+c^5=-15$, $a^5b^5+a^5c^5+b^5c^5=-198$, $a^5b^5c^5=-1$ and since $a+b+c=0$,
we obtain the starting equation.
My question is how we can solve this equation without previous way?
Thank you very much!

If $\alpha$ is a root of the polynomial $p(x)=x^3+15x^2-198x+1$ then $\alpha^{1/5}$ is a root of the polynomial $p(x^5)=(1-3x+x^3)\cdot q(x)$. If we manage to prove that $(1-3x+x^3)$ is the minimal polynomial of $\alpha^{1/5}$, the claim easily follows from Vieta's theorem. For the last part, it is simpler to go in the opposite direction.

Claim: if $\theta$ is a root of $x^3-3x+1$, then $\theta^5$ is a root of $x^3+15x^2-198x+1$.

Let we work in $\mathbb{Q}[x]/(x^3-3x+1)$. A base of this ring as a vector space over $\mathbb{Q}$ is given by $\{1,x,x^2\}$, and by computing a few polynomial remainders we have:
$$\begin{eqnarray*}1 &=& 1 \\ x^5 &=& -x^2+9x-3\\ x^{10} &=& 90x^2-109x+27\\ x^{15}&=&-1548 x^2+3417 x-1000 \end{eqnarray*} $$
so the previous Claim follows from gaussian elimination.
