Answers

2015-04-28T22:08:09+05:30
ax^3+bx^2+cx+d

From Vieta's formila we have
\alpha\beta+\beta\gamma+\gamma\alpha=\dfrac{c}{a}
\alpha\beta\gamma=-\dfrac{d}{a}

Now take the given expression
\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}\\~\\=\dfrac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma}\\~\\=\dfrac{\frac{c}{a}}{-\frac{d}{a}}\\~\\=-\frac{c}{d}
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