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.

We have
$$x^5 = -5x^3-1$$
and as t all roots are not $0$, the initial equation is equivalent to
$$1+5\frac{1}{x^2}+\frac{1}{x^5}=0 \Longleftrightarrow \frac{1}{x^5} = -\frac{5}{x^2}-1$$
Then, the sum is equal to
$$L:=-10-5\sum_{i=1}^5(x^3+\frac{1}{x^2})=-10-5\left(\sum_{\text{sym}}x_{\sigma(1)}^3+\sum_{\text{sym}}\frac{1}{x_{\sigma(1)}^2}\right)$$

For the first term $\sum_{\text{sym}}x_{\sigma(1)}^3$, using the Vieta's formula, we have:
$$\begin{align}
&\sum_{\text{sym}}x_{\sigma(1)} &= 0 \\
&\sum_{\text{sym}}x_{\sigma(1)}x_{\sigma(2)} &= 5\\
&\sum_{\text{sym}}x_{\sigma(1)}x_{\sigma(2)}x_{\sigma(3)} &= 0\\
\end{align}$$
We have from this   (normally, a real mathematician needs to know this formula by heart ;) )
$$\begin{align}
\sum_{\text{sym}}x_{\sigma(1)}^3&=\left(\sum_{\text{sym}}x_{\sigma(1)}  \right)^3 - 3\left(\sum_{\text{sym}}x_{\sigma(1)}\right)\left(\sum_{\text{sym}}x_{\sigma(1)}x_{\sigma(2)} \right)+3\left(\sum_{\text{sym}}x_{\sigma(1)}x_{\sigma(2)}x_{\sigma(3)}  \right)\\
& = 0^3 -3\cdot0\cdot5+3\cdot 0 \\
&= 0  \tag{1}
\end{align}$$

For the second term $\sum_{\text{sym}}\frac{1}{x_{\sigma(1)}^2}$, we remark that these $\frac{1}{x}$ are the roots of the equation
$$1 + 5y^2 + y^5 = 0 \tag{2}$$
Let denote $y_i$ the roots of $(2)$, we have $\sum_{\text{sym}}\frac{1}{x_{\sigma(1)}^2} = \sum_{\text{sym}}y_{\sigma(1)}^2$. By the same method, using the Vieta's formula, we have:
$$\begin{align}
&\sum_{\text{sym}}y_{\sigma(1)} &= 0 \\
&\sum_{\text{sym}}y_{\sigma(1)}y_{\sigma(2)} &= 0\\
\end{align}$$
and from that, we have
$$\begin{align}
\sum_{\text{sym}}y_{\sigma(1)}^2&=\left(\sum_{\text{sym}}y_{\sigma(1)}  \right)^2 - 2\left(\sum_{\text{sym}}y_{\sigma(1)}y_{\sigma(2)} \right)\\
& = 0^2 -2\cdot 0 \\
&= 0  \tag{3}
\end{align}$$

So, we can compute $L$ from $(1)$ and $(3)$
$$\color{red}{L} = -10 -5 \left( 0 + 0   \right) \color{red}{= -10}$$
