$\alpha,\beta,\gamma$ are the roots of the equation $x^3 − 9x + 9 = 0$. Find the value of $ \alpha^{-5}+\beta^{-5}+\gamma^{-5}$ I've simplified the expression to get
$$\frac{(\alpha\beta)^5+(\beta\gamma)^5+(\gamma\alpha)^5}{(\alpha\beta\gamma)^5}.$$
Now all I need to find is $\sum (\alpha\beta)^5$ given that $\sum \alpha\beta=-9$ (by Vieta's relations).
If it helps, I found that:
$$\sum (\alpha\beta)^2=81,$$ and
$$\sum (\alpha\beta)^3=-486.$$

Let $a=\alpha^{-1}, b=\beta^{-1},c=\gamma^{-1}.$ Then, we have $a+b+c=\alpha^{-1}+\beta^{-1}+\gamma^{-1}=(\alpha\beta+\beta\gamma+\gamma\alpha)/(\alpha\beta\gamma)=1,$ $abc=(\alpha\beta\gamma)^{-1}=-1/9,$ and $ab+bc+ca=(\alpha+\beta+\gamma)/(\alpha\beta\gamma)=0.$ Then, we have
\begin{equation}
\begin{split}
\alpha^{-5}+\beta^{-5}+\gamma^{-5}&=a^5+b^5+c^5\\
&=(a+b+c)^5-5(ab+bc+ca)(a+b+c)^3\\
&\ \ \ \ \ +5abc(a+b+c)^2+5(ab+bc+ca)^2(a+b+c)\\
&\ \ \ \ \ -5abc(ab+bc+ca),
\end{split}
\end{equation}
and all that is left to do is substitute.
P.S. The equation above can be obtained by repeated use of Newton's identity, which can be used to express $a^i+b^i+c^i$ using the basic symmetric polynomials $a+b+c,ab+bc+ca,abc.$
