The correct option is option (a) $\frac{-b}{a}$
$\alpha, 0$ and 0 be the zeroes of $a x^{3}+b x^{2}+c x+d=0$
Then the sum of zeroes $=\frac{-b}{a}$
$\Rightarrow \alpha+0+0=\frac{-b}{a}$
$\Rightarrow \alpha=\frac{-b}{a}$
Hence, the third zero is $\frac{-b}{a}$.