If two zeroes of the polynomial x3 + x2 − 9x − 9 are 3 and −3,

Let $\alpha=3$ and $\beta=-3$ be the given zeros and $\gamma$ be the third zero of the polynomial $x^{3}+x^{2}-9 x-9$ then

By using $\alpha+\beta+\gamma=\frac{-\text { Coefficient of } x^{2}}{\text { Coefficient of } x^{3}}$

$\alpha+\beta+\gamma=\frac{-1}{1}$

$\alpha+\beta+\gamma=-1$

Substituting $\alpha=3$ and $\beta=-3$ in $\alpha+\beta+\gamma=-1$, we get

$3-3+\gamma=-1$

$\gamma=-1$

Hence, the correct choice is $(a)$