Express the sum of three roots as combination of quotients I'm trying to do an exercise relative to symmetrical polynomials. 
We're given the following polynomial:
$ X^3 + pX +q = 0$
With $x_1, x_2, x_3 $ its roots.
We're asked to give an expression with $p, q$ of the following sum :
$x_1^8 + x_2^8 + x_3^8 $
The obvious way to do this is way to messy, but I recall our teacher saying there was a trick when working with high exponents like these ones.
Please notice that this expression is symmetric, so we can use Cardano-Vieta to solve this. 
Thank you for reading.
 A: I usually use this trick. Let's call:
$$S_j=x_1^j+x_2^j+x_3^j$$
We are searching $S_8$. Trivially:
$$\left\{\begin{matrix}
x_1^3+px_1+q=0\\ 
x_2^3+px_2+q=0\\ 
x_3^3+px_3+q=0
\end{matrix}\right.$$
If we multiply both sides for the monomial $x^j$ the sum of the powers of the roots doesn't change cause we are adding $n$ roots that are equal 0. So:
$$\left\{\begin{matrix}
x_1^{j+3}+px_1^{j+1}+qx_1^{j}=0\\ 
x_2^{j+3}+px_2^{j+1}+qx_1^{j}=0\\ 
x_3^{j+3}+px_3^{j+1}+qx_1^{j}=0
\end{matrix}\right.$$
Sum all of the terms and you get:
$$S_{j+3}=-pS_{j+1}-qS_{j}$$
So now we must calculate $S_2$ and $S_3$  and apply the recursion some times. $S_2$ is a standard calculus:
$$S_2=x_1^2+x_2^2+x_3^2=(x_1+x_2+x_3)^2-2(x_1x_2+x_1x_3+x_2x_3)$$
Applying Vieta formulas:
$$x_1+x_2+x_3=0$$
$$x_1x_2+x_1x_3+x_2x_3=p$$
So:
$$S_2=-2p$$
Moreover for the recursion:
$$S_{3}=-pS_{1}-qS_{0}=-3q$$
And for the recursive formula:
$$S_5=-pS_3-qS_2=3pq+2pq=5pq$$
$$S_8=-pS_6-qS_5$$
A last effort for $S_7$:
$$S_4=-pS_2-qS_1=2p^2$$
$$S_6=-pS_4-qS_3=-2p^3+3q^2$$
And finally:
$$S_8=-p(-2p^3+3q^2)-q(5pq)$$
$$S_8=2p^4-8pq^2 $$
:)
A: Hint:
Denote $\;\sigma_1=\sum_i x_i$, $\;\sigma_2=\sum_{i\ne j}x_ix_j$, $\; \sigma_3=x_1x_2x_3$. From the form of the equation, we know that 
$$\sigma_1=0,\enspace \sigma_2=p,\quad \sigma_3=-q.$$
Now rewrite the equation as  $X^3=-pX-q$. We deduce
\begin{cases}
x_i^3=-px_i-q, \quad\text{ hence} \\
x_i^6=(px_i+q)^2=p^2x_i^2 +2pqx_i+q^2 \\
x_i^8=p^2x_i^4 +2pqx_i^3+q^2x_i^2 =\dotsm
\end{cases}
Can you continue to express $x_i^8$ as a quadratic polynomial in $x_i$, and, from there, express $\sum_i x_i^8$ with $\sigma_1, \sigma_2,\sigma_3$?
A: Use Newton's Identities (to relate sums of powers of roots to coefficients of the polynomial).
Using the notation at the cited page, you want $p_8(x_1,x_2,x_3)$, where (henceforth suppressing "$(x_1,x_2,x_3)$") 
$$  e_0 = 1 , e_1 = 0 , e_2 = p , e_3 = -q  \text{, and } e_{\geq 4} = 0  \text{.}  $$
Then\begin{align*}
p_1 &= e_1 = 0  \\
p_2 &= e_1 p_1 - 2 e_2  \\
    &= 0 \cdot 0 - 2 \cdot p = -2 p  \\
p_3 &= e_1 p_2 - e_2 p_1 + 3 e_3  \\
    &= 0 - 0 + 3 (-q)  = -3 q  \\
p_4 &= e_1 p_3 - e_2 p_2 + e_3 p_1 - 0  \\
    &= 0 - p(-2p) + 0 - 0 = 2 p^2  \\
    &\vdots
\end{align*}

 \begin{align*}  p_5 &= e_1 p_4 - e_2 p_3 + e_3 p_2 - 0  \\  &= 0 - p(-3q) - q(-2p) = 5 pq  \\  p_6 &= e_1 p_5 - e_2 p_4 + e_3 p_3 - 0  \\  &= 0 - p(2p^2) - q(-3q) = -2 p^3 + 3 q^2  \\  p_7 &= e_1 p_6 - e_2 p_5 + e_3 p_4 - 0  \\  &= 0 - p(5pq) - q(2p^2) = -7 p^2 q  \\  p_8 &= e_1 p_7 - e_2 p_6 + e_3 p_5 - 0  \\  &= 0 - p(-2 p^3 + 3 q^2) - q(5 pq)  \\  &= 2 p^4 - 8 p q^2  \text{.}  \end{align*} 

Note that, since $e_{\geq 4} = 0$, once we get to $p_4$ and forever after, we are performing the same dot product with the previous three terms, so this can be rolled up into a matrix multiplication and made much faster, especially for $p_\text{huge number}$.

An entirely different way to go is this.  We want to reduce $x_1^8 + x_2^8 +x_3^8$ by the relations $x_1^3 + p x_1+ q = 0$, $x_2^3 + p x_2+ q = 0$, $x_3^3 + p x_3+ q = 0$, $x_1 + x_2 + x_3 = 0$, $x_1 x_2 + x_1 x_3 + x_2 x_3 = p$, and $x_1 x_2 x_3 = -q$.  Constructing a Groebner basis of those relations with variable order $x_1, x_2, x_3$, gives $$ \{x_3^3 + p x_3 + q, x_2^2 + x_2 x_3 + x_3^2 + p, x_1 + x_2 + x_3\}  $$
and reducing $x_1^8 + x_2^8 + x_3^8$ by this basis gives 

 $$  2p^4 - 8 p q^2  \text{.}  $$ 

We can make Wolfram Alpha do all the tedious work (and use a shorter basis specification since the relations are redundant).  The result is the last expression, the (canonical) remainder on dividing $x_1^8 + x_2^8 + x_3^8$ by the relations.
A: Denoting the polynomial by $P(X)$, we can consider the expansion of $\log\left[P(X)/X^3\right]$ in powers of $X^{-1}$. We have:
$$\frac{P(X)}{X^3} =\prod_{k=1}^3\left(1-\frac{x_k}{x}\right)$$
Therefore:
$$\log\left[\frac{P(X)}{X^3}\right] =-\sum_{r=1}^\infty\frac{S_r}{rX^r}$$
where
$$S_r = \sum_{k=1}^3 x_k^r$$
So, we need to compute the series expansion of 
$$\log\left(1 + \frac{p}{X^2} + \frac{q}{X^3}\right)$$
the sum of the 8th powers of the roots is then the coefficient of $X^{-8}$ times $-8$. In this particular case, extracting this coefficient requires very little effort as the only contributions come from the third and fourth power of $\frac{p}{X^2} + \frac{q}{X^3}$. The fourth power yields a contribution
$$(-8)\times \left(-\frac{1}{4}\right) p^4= 2 p^4$$
The third power yields a contribution via the square of the second term multiplied by the first, this is
$$(-8)\times \frac{1}{3} \times 3 p q^2=-8 p q^2$$ 
The result is therefore:
$$S_8 = 2 p^4 - 8 p q^2$$
