The problem states, if $\alpha, \beta, \gamma$ are root of polynomial $ f(z) =z^3-z-1 $ then value of integral,
$ \frac{1}{2\pi\iota}\int_C{\bigl(z^9 \frac{f'(z)}{f(z)}\bigr)} dz $
where c is curve containing all zeros of $f(z)$
What i know about this problem: if we know zeros of $f(z)$ then we can solve it using argument theorem which states this integral = $\alpha^9 + \beta ^9 + \gamma^9$, since $f(z)$ doesn't have any pole. but finding pole of this function is not easy. How to approach such questions.
EDIT: I did tried some manipulation but no luck. $ \alpha$ is zero of $f(z)$ then $\alpha^3-\alpha-1=0 $ or $\alpha^3=\alpha+1 $
similarly $\beta^3=\beta+1 $ and $\gamma^3=\gamma+1 $
so $\alpha^9 + \beta ^9 + \gamma^9$= $ (\alpha+1)^3+(\beta+1)^3+(\gamma+1)^3$ and then proceeding but no luck
**EDIT-2:**Continuing the above way,$\alpha^9 + \beta ^9 + \gamma^9$=$\alpha^3+1+3\alpha^2+3\alpha + \beta^3+1+3\beta^2+3\beta + \gamma^3+1+3\gamma^2+3\gamma$ =
$( \alpha +1) +1 +3\alpha^2+3\alpha +( \beta+1) +1 +3\beta^2+3\beta + ( \gamma+1) +1 +3\gamma^2+3\gamma $ $\tag{1}\label{eq1}$ Now using the fact that sum of zeros of polynomial is coefficient second highest power. so $\alpha + \beta +\gamma = 0$ and sum of product of roots $\alpha\beta +\alpha\gamma+\beta\gamma = -1$
(source for this information:https://www.myqbook.com/MathConcept/597/The-sum-and-product-of-the-roots-of-a-cubic-equation)
squaring both sides, $\alpha^2 +\beta^2 +\gamma^2 +2\alpha\beta +2\alpha\gamma +2\gamma\beta=0$ =
$\alpha^2 +\beta^2 +\gamma^2 =-2(\alpha\beta +\alpha\gamma +\gamma\beta)=2$
putting these values in equation 1 $4(\alpha +\beta+\gamma + 6 +3(\alpha^2+\beta^2 +\gamma^2) =6+3*2=12 $
Obviously solution given in comment by @conrad is much more elegant.