A) \[\frac{c}{a}\]
B) \[\frac{a}{c}\]
C) \[-\frac{a}{c}\]
D) \[-\frac{c}{a}\]
Correct Answer: B
Solution :
| [b] Since \[\alpha ,\beta ,\gamma \] are the zeroes of the polynomial |
| \[f(x)={{x}^{3}}-a{{x}^{2}}+bx-c.\] |
| \[\therefore \,\,\,\,\,\alpha +\beta +\gamma =-\frac{(-a)}{1}=a,\] \[\alpha \beta \gamma =-\frac{(-c)}{1}=c\] |
| Now, \[\frac{1}{\alpha \beta }+\frac{1}{\beta \gamma }+\frac{1}{\gamma \alpha }=\frac{\alpha +\beta +\gamma }{\alpha \beta \gamma }=\frac{a}{c}\] |
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