<s>[INST] Evaluating $\sum_{cyc}\frac{ab}{c^5}$, where $a$, $b$, $c$ are the roots of $x^3-px+1=0$. How to avoid a large expansion? 
Given the polynomial $x^{3}-p{x}+1=0$, evaluate $\frac{bc}{a^{5}}+\frac{ac}{b^{5}}+\frac{ab}{c^{5}}$ in terms of $p$ if $a,b$ and $c$ are the roots of the polynomial.

My attempt involved:
$$\frac{bc}{a^{5}}+\frac{ac}{b^{5}}+\frac{ab}{c^{5}}=\frac{\left(ab+ac+bc\right)^{6}-4...}{\left(-1\right)^{5}}$$
but the power $6$ is just ridiculous. There must be a way to evaluate this without having to expand a trinomial to the power of $6$ by hand. Any ideas?
 [/INST] As Martin R noted we have to find the value of $$t_6=p^6+q^6+r^6$$ where  $p.q.r$ are the roots of $x^3-px^2+1$. Now let $t_k=p^k+q^k+r^k$ for $k\ge 3$ indeed from $x^3=px^2-1$ we get the recurrence $$t_k=pt_{k-1}-3$$ Now we have to find $t_6$ when we know the value of $t_2=p^2$....
The rest should be easy
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