Find the sum of $\sum_{i=1}^{5}x^5_i+\sum_{i=1}^{5}\frac{1}{x^5_i}$ 
Suppose $x^5+5x^3+1=0$ and $x_i$ denotes all the complex roots. Find the sum of
$$\sum_{i=1}^{5}x^5_i+\sum_{i=1}^{5}\frac{1}{x^5_i}$$

This polynomial is irreducible over $\Bbb Q[x]$.
I used Vieta's general formulas, however I am not sure if my calculations correct. Here are my calculations.
$$x_1x_2+x_1x_3+x_1x_4+x_1x_5+x_2x_3+x_2x_4+x_2x_5+x_3x_4+x_3x_5+x_4x_5=5$$
$$x_1x_2x_3x_4x_5=-1$$
$$x_1+x_2+x_3+x_4+x_5=0$$
$$x_1x_2x_3+x_1x_2x_4+x_1x_2x_5+x_1x_3x_4+x_1x_3x_5+x_1x_4x_5+x_2x_3x_4+x_2x_3x_5+x_2x_4x_5+x_3x_4x_5=0$$
$$x_1x_2x_3x_4+x_1x_2x_3x_5+x_1x_2x_4x_5+x_1x_3x_4x_5+x_2x_3x_4x_5=0$$
This seems so difficult using these formulas. I couldn't make any solution.
How can I simplify this summation?
$$x_1^5+x_2^5+x_3^5+x_4^5+x_5^5$$
I've tried bring the fractions to common denominator and I got
$$\frac {1}{x_1^5}+\frac {1}{x_2^5}+\frac {1}{x_3^5}+\frac {1}{x_4^5}+\frac {1}{x_5^5}=\frac {(x_1x_2x_3x_4)^5+(x_1x_2x_3x_5)^5+(x_1x_2x_4x_5)^5+(x_1x_3x_4x_5)^5+(x_2x_3x_4x_5)^5}{(x_1x_2x_3x_4x_5)^5}=-(x_1x_2x_3x_4)^5-(x_1x_2x_3x_5)^5-(x_1x_2x_4x_5)^5-(x_1x_3x_4x_5)^5-(x_2x_3x_4x_5)^5$$
But, I couldn't make any further progress from here.

According to Fundamental Theorem of Symmetric Polynomials,
symmetric polynomials can expressed as polynomial of elementary symmetric polynomials, i.e. the LHS of Vieta's formula.
Then by using math software, the calculation is as follow.
The result of $\sum_{i=1}^{5} x_i^5$ is
$$
s_{1}^{5} - 5 s_{1}^{3} s_{2} + 5 s_{1}^{2} s_{3} + 5 s_{1} s_{2}^{2} - 5 s_{1} s_{4} - 5 s_{2} s_{3} + 5 s_{5}$$
The result of $\sum_{i=1}^{5} \frac{1}{x_i^5}$ is
$$ \frac{- 5 s_{1} s_{4} s_{5}^{3} - 5 s_{2} s_{3} s_{5}^{3} + 5 s_{2} s_{4}^{2} s_{5}^{2} + 5 s_{3}^{2} s_{4} s_{5}^{2} - 5 s_{3} s_{4}^{3} s_{5} + s_{4}^{5} + 5 s_{5}^{4}}{s_{5}^{5}}$$
Given the equation $x^5+5x^3+1=0$,
$s_1=0,s_2=5,s_3=0,s_4=0,s_5=-1$.
Hence the final result is $-10$.
