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?
 A: 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, $$\begin{align*} \sum_{i=1}^5 r_i^{11} &= \sum_{i=1}^5 (r_i - 125r_i^{-1} + 10r^2 - 25r^{-2}) \\ &= 0 - 125 (-5) + 10(0) - 25(25) = 625 - 625 \\ &= 0. \end{align*}$$
A: You can use the technique of Newton's Sums which @user157227's comment hinted at
SOLUTION:
Basically imagine if you factored the polynomial into its 5 roots
$$(x - r_1)(x - r_2)..(x - r_5)$$
And then expanded that out:
it quickly will become evident you get an answer of the form
$$a_0 + a_1x + a_2x^2 ... a_4x^4 + x^5$$
such that
$$ \begin{matrix} a_4 = r_1 + r_2  ... r_5 \\ a_3  = r_1r_2  + r_1r_3 + ... + r_1r_5  + r_2 r_3  + ... + r_2 r_5 + ... r_4 r_5 \\ a_2 = r_1 r_2 r_3  + r_1 r_2 r_4 + ... r_3 r_4 r_5 \\ a_1 = r_1r_2r_3r_4 + ... +r_2 r_3r_4r_5 \\ a_0 = r_1r_2r_3r_4r_5 \end{matrix} $$
So to solve the case for sum of all 11th powers of the roots we need to only solve for the third, 2nd and single powers based on your formula. We start with:
the sum of all first powers is merely $a_4: \ (r_1 + r_2 .. r_5)$
To find the square of all the powers note that we can compute
$$(r_1 + r_2  + \ ... \ + r_5)^2 = \sum(r_i^2) + 2\sum(r_ir_j) = \sum(r_i^2) + 2a_3 \rightarrow (a_4)^2 - 2a_3 = \sum(r_i^2) = \Omega_1 $$
To find the cube of all the powers 
$$(r_1 + r_2 + \ ... \ r_5)^3 = \sum(r_i)^3 + 3\sum(r_i^2r_j) + 6\sum(r_ir_jr_k) = 6(r_1r_2r_3 + ... r_3r_4r_5) + 3(r_1^2 + ... + r_5^2)(r_1 + ... + r_5) - 2\sum(r_i)^3 \rightarrow \sum(r_i)^3 = -\frac{1}{2} \left( a_4^3  - 6(a_2) + 3(\Omega_1)a_4 \right) = \Omega_2$$
Now since $a_4,a_3,a_2$ are equal to 0 it follows that
$$25\sum(\Omega_2) + 10\sum(\Omega_1) + a_4 = 0$$
A: You have made a good start.  Now remember that the coefficient of $x^4$ in the original equation is the negative of the sum of the roots.  You have an unfortunate reuse of $\alpha$, once as one of the roots and once as a generic root, but in the last equation $\sum \alpha=0$  What does this mean for the other terms?
