If $\alpha, \beta, y$ are are the zeros of the polynomial $f(x)=x^{3}-p x^{2}+q x-r$, then $\frac{1}{\alpha \beta}+\frac{1}{\beta \gamma}+\frac{1}{\gamma \alpha}=$
We have to find the value of $\frac{1}{\alpha \beta}+\frac{1}{\beta \gamma}+\frac{1}{\gamma \alpha}$
Given $\alpha, \beta, \gamma$ be the zeros of the polynomial $f(x)=x^{3}-p x^{2}+q x-r$
$\alpha+\beta+\gamma=\frac{-\text { Coefficient of } x^{2}}{\text { Coefficient of } x^{3}}$
$=\frac{-(p)}{1}$
$=p$
$\alpha \beta \gamma=\frac{-\text { Constant term }}{\text { Coefficient of } x^{3}}$
$=\frac{-(r)}{1}$
$=r$
Now we calculate the expression
$\frac{1}{\alpha \beta}+\frac{1}{\beta \gamma}+\frac{1}{\gamma \alpha}=\frac{\gamma}{\alpha \beta \gamma}+\frac{\alpha}{\alpha \beta \gamma}+\frac{\beta}{\alpha \beta \gamma}$
$\frac{1}{\alpha \beta}+\frac{1}{\beta \gamma}+\frac{1}{\gamma \alpha}=\frac{\alpha+\gamma+\beta}{\alpha \beta \gamma}$
$\frac{1}{\alpha \beta}+\frac{1}{\beta \gamma}+\frac{1}{\gamma \alpha}=\frac{p}{r}$
Hence, the correct choice is (b)