Sum of the $11^\mathrm{th}$ power of the roots of the equation $x^5+5x+1=0$ 
Find the sum of the $11^\mathrm{th}$ power of the all roots of the equation
$$
x^5+5x+1=0
$$

My Attempt:
Let $R=\{\alpha,\beta,\gamma,\delta,\mu\}$ be the set of all roots of the equation ${x^5+5x+1=0}$, and let $x\in R$. Then we have
$$
\begin{align}
x^5 &= -\left(5x+1\right)\\
x^{10}&=25x^2+1+10x\\
\tag1 x^{11}&=25x^3+10x^2+x
\end{align}
$$
Taking the sum of $(1)$ on all elements $x\in R$ gives us
$$
\sum_{x\in R}x^{11}=25\cdot\sum_{x\in R}x^3+10\cdot\sum_{x\in R}x^2+\sum_{x\in R}x
$$
How can I solve the problem from this point?

If $r$ is a root, then $r^5 = -5r - 1$, hence $r^{10} = (5r + 1)^2$ and $r^{11} = r(5r + 1)^2$.  This greatly simplifies subsequent calculations.  We can also note $r^3 = -5r^{-1} - r^{-2}$, so we get $$\begin{align*} r^{11} &= 25r^3 + 10r^2 + r \\ &= 25(-5r^{-1} - r^{-2}) + 10r^2 + r \\ &= r - 125r^{-1} + 10r^2 - 25r^{-2}\end{align*}.$$  So if $\{r_i\}_{i=1}^5$ are the five nonzero* complex roots of the given quintic $$f(x) = (x-r_1)(x-r_2)(x-r_3)(x-r_4)(x-r_5) = x^5 + 5x + 1,$$ then $\{r_i^{-1}\}_{i=1}^5$ are roots of the polynomial $$\begin{align*} f(1/x)x^5 &= x^5 \prod_{i=1}^5 (x^{-1} - r_i) = \prod_{i=1}^5 r_i (r_i^{-1} - x) \\ &= -r_1 r_2 r_3 r_4 r_5 \prod_{i=1}^5 (x - r_i^{-1}) \\ &= x^5 + 5x^4 + 1. \end{align*}$$  (* We know they are nonzero since $f(0) = 1$.)
Next, how do we find a polynomial whose roots are $\{r_i^2\}_{i=1}^5$?  This is easily done by observing that $$\prod_{i=1}^5 (x - r_i^2) = \prod_{i=1}^5 (\sqrt{x} - r_i)(\sqrt{x} + r_i) = -f(\sqrt{x})f(-\sqrt{x}) = x^5 + 10x^3 + 25x - 1.$$  And a polynomial with roots $\{r_i^{-2}\}_{i=1}^5$ is, by a combination of the above reasoning, $$x^5 - 25x^4 - 10x^2 - 1.$$
Consequently, $$\begin{align*} \sum_{i=1}^5 r_i^{11} &= \sum_{i=1}^5 (r_i - 125r_i^{-1} + 10r^2 - 25r^{-2}) \\ &= 0 - 125 (-5) + 10(0) - 25(25) = 625 - 625 \\ &= 0. \end{align*}$$
