$\alpha,\beta,\gamma$ 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
$$\sum (\alpha\beta)^3=-486.$$
 A: You can express $\sum(\alpha\beta)^5$ in terms of the elementary symmetric polynomials by the standard method, by repeatedly subtracting the first monomial in the lexicographical order. Here's the first step:
$$\sum(\alpha\beta)^5-\left(\sum(\alpha\beta)\right)^5=5\sum\alpha^5\beta^4\gamma+10\sum\alpha^5\beta^3\gamma^2+20\sum\alpha^4\beta^4\gamma^2+30\sum\alpha^4\beta^3\gamma^3.$$
The next step would be to subtract $5(\sum\alpha)^{5-4}(\sum\alpha\beta)^{4-1}(\sum\alpha\beta\gamma)^{1-0}$. This is a bit tedious, but you should eventually find a polynomial in the symmetric sums consisting of $7$ monomials.
Alternatively, you can look up Newton's identities and plug in your values.
A: Thanks to everyone who solved it, but I've solved it myself now.
Let $a=\alpha\beta, b=\beta\gamma, c=\gamma\alpha$.
a,b,c are the roots of $x^3+9x^2-81=0$
We know that
$$a(a^3+9a^2-81)=0$$
$$a^4+9a^3-81a=0$$      (1)
$$b^4+9b^3-81b=0$$      (2)
$$c^4+9c^3-81c=0$$      (3)
Adding (1), (2) and (3)
$$\sum a^4 + 9\sum a^3 -81\sum a=0$$
$$\sum a^4 = -9\sum a^3 +81\sum a=3645$$
Similarly we get,
$$\sum a^5= -9\sum a^4 +81\sum a^2=-26244$$
Now,
$$\alpha^{-5}+\beta^{-5}+\gamma^{-5} = \frac{\sum a^5}{-9^5}$$
$$=\frac{-26244}{-9^5}=\frac{4}{9}$$
A: Here is a quite quick solution using linear recurrences.
Replacing $x=\frac 1y$  in $x^3 − 9x + 9 = 0$ we conclude 
$$\frac 1{\alpha},\frac 1{\beta},\frac 1{\gamma} \text{ are the roots of }y^3-y^2+\frac 19=0$$
So, $$\frac 1{\alpha^{-5}}+\frac 1{\beta^{-5}}+\frac 1{\gamma^{-5}} = a_5$$
in the recursion
$$a_{n+3} = a_{n+2}-\frac 19 a_n \text{ with } a_0 = \frac 1{\alpha^{0}}+\frac 1{\beta^{0}}+\frac 1{\gamma^{0}}=1$$
Vieta gives the other starting values
$$a_1 = 1, a_2 = a_1^2 - 2\cdot 0 = 1$$
Now, 
$$a_3 =1-\frac 13 = \frac 23,\; a_4 = \frac 23 - \frac 19= \frac 59$$
and finally
$$a_5 = \frac 59 - \frac 19 = \boxed{\frac 49}$$
A: 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 $a+b+c,ab+bc+ca,abc.$
