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
$$\sum (\alpha\beta)^3=-486.$$
Best Answer
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.