If $\alpha, \beta$ and $\gamma$ are roots of the equation $x^3+4 x-19=0$. Then, the value of $\frac{\alpha^3}{19-4 \alpha}+\frac{\beta^3}{19-4 \beta}+\frac{\gamma^3}{19-4 \gamma}$ is equal to

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Aicharya · Accepted Answer

$\alpha, \beta$ and $\gamma$ are roots of $x^3+4 x-19=0$ Then, $\alpha+\beta+\gamma=0$ $\begin{aligned} \alpha \beta+\beta \gamma+\gamma \alpha & =4 \ \alpha \beta \gamma & =19\end{aligned}$ $ \alpha \beta \gamma=19 $ From Eq. (ii), $\alpha(\beta+\gamma)+\beta \gamma=4$ $ \begin{gathered} \alpha(-\alpha)+\beta \gamma=4 \ \Rightarrow-\alpha^2+\frac{19}{\alpha}=4 \ -\alpha^3+19=4 \alpha \end{gathered} $ [using Eq. (i)] [using Eq. (iii)] or $ \alpha^3=19-4 \alpha $ This gives $\frac{\alpha^3}{19-4 \alpha}=1$ Similarly, from Eq. (ii), $ \begin{aligned} & \beta(\alpha+\gamma)+\gamma \alpha=4 \ & \Rightarrow \quad \beta(-\beta)+\gamma \alpha=4 \ & \Rightarrow \quad-\beta^2+\frac{19}{\beta}=4 \ & \Rightarrow \quad \beta^3=19-4 \beta \ & \Rightarrow \quad \frac{\beta^3}{19-4 \beta}=1 \ & \end{aligned} $ And, from Eq. (ii) $ \begin{array}{r} \gamma(\alpha+\beta)+\alpha \beta=4 \ \gamma(-\gamma)+\frac{19}{\gamma}=4 \end{array} $ $ \Rightarrow \quad \frac{\gamma^3}{19-4 \gamma}=1 $ Adding Eqs.(iv), (v) and (vi), we get $ \frac{\alpha^3}{19-4 \alpha}+\frac{\beta^3}{19-4 \beta}+\frac{\gamma^3}{19-4 \gamma}=3 $

If $\alpha, \beta$ and $\gamma$ are roots of the equation $x^3+4 x-19=0$. Then, the value of…

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