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.
 A: 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 $(3)$
$$\color{red}{L} = -10 -5 \left( 0 + 0   \right) \color{red}{= -10}$$
A: Let $$S_k=\sum_{i=1}^5x_i^k$$ and then we know that $S_1=0$. Next from the given equation we have $$S_5+5S_3+5=0\tag{1}$$ and hence we need to evaluate $S_3$ first. To that end we note that the roots are non-zero and then we can divide the given polynomial equation by $x^2$ to get $$x^3+5x+x^{-2}=0$$ which leads to $$S_3+5S_1+S_{-2}=0$$ ie $S_3=-S_{-2}$. Now $S_{-k} =\sum_{i=1}^5y_i^k$ where $y_i=1/x_i$ are roots of $$1/y^5+5/y^3+1=0$$ ie $$y^5+5y^2+1=0\tag{2}$$ Then we have $$S_{-2}=S_{-1}^2-2\sum y_iy_j=0$$ and thus we get $S_3=0$. Using $(1)$ we get $S_5=-5$.
To evaluate $S_{-5}$ we use equation $(2)$ to get $$S_{-5}+5S_{-2}+5=0$$ and using $S_{-2}=0$ we get $S_{-5}=-5$ and thus $S_{5}+S_{-5}=-10$.
A: if you have a polinomial of any degree, e.g $p(x)=ax^5+bx^3+c$, with roots $r_1, r_2, \cdots, r_5$, lets denote $S_k = \sum_{i=1}^{5}r_i^k = r_1^k+r_2^k+\cdots+r_5^k$.2
This is a fact well known. But the additional information of this answer is that for any $k\in\mathbb{C}$ is true that:
$$aS_{K+5}+bS_{K+3}+cS_K=0$$
So, if you known $S_0, S_1, S_2...$ you can use it to known the rest of the values that you require. For example: $aS_5 +bS_3 + cS_1 = 0$ in the example showed.
Also, you can have in a easy manner the values of $S_{-1}, S_{-2}, S_{-3}, \cdots$
you have to make a "non-algebraic division$^1$" $p'(x)/p(x)$ with Horner-Method and the coefficients will be the values of $S_0, S_{-1}, S_{-2}, S_{-3}, ...$.
For example. for your polinomial in the example: (if anyone can edit on latex would be better)

Note that this method works for any polinomial and any value of $k$
A: 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 $-10$.
A: Since $x=0$ is not one of the roots of $x^5+5x^3+1=0$, then you can divide all terms of the polynomial by $x^3:$
$$x^2+\frac {1}{x^3}=-5$$
So, why did I do this? Because, we want to construct an “useful relationship” between the roots. Then, with a little observation, we want to square both side:
$$x^4+\frac 2x+\frac {1}{x^6}=25$$
Multiplying both side by $x$, we have:
$$x^5+\frac {1}{x^5}=25x-2$$
Finally, we obtain
$$
\begin{align}\sum_{1\leqslant i\leqslant 5}x_i^5+\sum_{1\leqslant i\leqslant 5}\frac {1}{x_i^5}&=\sum_{1\leqslant i\leqslant 5}\left(x_i^5+\frac{1}{x_i^5}\right)\\
&=\sum_{1\leqslant i\leqslant5}\left(25x_i-2\right)\\
&=25\cdot 0-2\cdot 5\\
&=-10.\end{align}
$$
and you are done.
