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If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^3+3 x^2-7 x+5=0$, then the value of $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$ is
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The correct answer is:
$\frac{7}{5}$
Given, cubic equation,
$$
x^3+3 x^2-7 x+5=0 ...(i)
$$
Let $\alpha, \beta$ and $\gamma$ are the roots of Eq. (i)
$$
\begin{array}{ll}
\therefore \quad & \alpha+\beta+\gamma=\frac{-3}{1}=-3 \\
& \alpha \beta+\beta \gamma+\gamma \alpha=\frac{-7}{1}=-7 \\
& \alpha \beta \gamma=\frac{-5}{1}=-5 \\
\therefore \quad & \frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=\frac{\beta \gamma+\alpha \gamma+\alpha \beta}{\alpha \beta \gamma}=\frac{-7}{-5}=\frac{7}{5}
\end{array}
$$
$$
x^3+3 x^2-7 x+5=0 ...(i)
$$
Let $\alpha, \beta$ and $\gamma$ are the roots of Eq. (i)
$$
\begin{array}{ll}
\therefore \quad & \alpha+\beta+\gamma=\frac{-3}{1}=-3 \\
& \alpha \beta+\beta \gamma+\gamma \alpha=\frac{-7}{1}=-7 \\
& \alpha \beta \gamma=\frac{-5}{1}=-5 \\
\therefore \quad & \frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=\frac{\beta \gamma+\alpha \gamma+\alpha \beta}{\alpha \beta \gamma}=\frac{-7}{-5}=\frac{7}{5}
\end{array}
$$
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