If $\alpha, \beta, y$ are the zeros of the polynomial $f(x)=a x^{3}+b x^{2}+c x+d$, then $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=$
(a) $-\frac{b}{d}$
(b) $\frac{c}{d}$
(c) $-\frac{c}{d}$
(d) $-\frac{c}{a}$
We have to find the value of $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$
Given $\alpha, \beta, \gamma$ be the zeros of the polynomial $f(x)=a x^{3}+b x^{2}+c x+d$
We know that
$\alpha \beta+\beta \gamma+\gamma \alpha=\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{3}}$
$=\frac{c}{a}$
$\alpha \beta \gamma=\frac{-\text { Constant term }}{\text { Coefficient of } x^{3}}$
$=\frac{-d}{a}$
So
$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=\frac{\beta \gamma+\alpha \gamma+\alpha \beta}{\alpha \beta \gamma}$
$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=\frac{\frac{c}{a}}{-\frac{d}{a}}$
$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=\frac{c}{a} \times\left(-\frac{a}{d}\right)$
$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=-\frac{c}{d}$
Hence, the correct choice is (c)