Question

Let $\alpha, \beta, \gamma$ be the roots of $x^{3}+x+10=0$ and
$\alpha_{1}=\frac{\alpha+\beta}{\gamma^{2}}, \beta_{1}=\frac{\beta+\gamma}{\alpha^{2}}, \gamma_{1}=\frac{\gamma+\alpha}{\beta^{2}} .$ Then, the
value of $\left(\alpha_{1}^{3}+\beta_{1}^{3}+\gamma_{1}^{3}\right)-\frac{1}{10}\left(\alpha_{1}^{2}+\beta_{1}^{2}+\gamma_{1}^{2}\right)$ is

• A. $\frac{1}{10}$
• B. $\frac{1}{5}$
• C. $\frac{3}{10}$
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

Since, $\alpha, \beta, \gamma$ are the roots of the equation
$x^{3}+x+10=0$
$\therefore \alpha+\beta+\gamma=0 \,......(i)$
Now, $\alpha_{1}=\frac{\alpha+\beta}{\gamma^{2}}=\frac{-\gamma}{\gamma^{2}}=\frac{-1}{\gamma}$
Similarly, $\beta_{1}=\frac{-1}{\alpha}$ and $\gamma_{1}=\frac{-1}{\beta}$
$\therefore \alpha_{1}, \beta_{1}, \gamma_{1}$ are the roots of equation $f\left(\frac{-1}{x}\right)=0$.
Now, $f\left(\frac{-1}{x}\right)$ will be $\left(-\frac{1}{x}\right)^{3}+\left(-\frac{1}{x}\right)+10=0$
$\Rightarrow 10 x^{3}-x^{2}-1=0$
$\Rightarrow 10 \alpha_{1}^{3}-\alpha_{1}^{2}-1=0$
$\Rightarrow \alpha_{1}^{3}=\frac{1}{10} \alpha_{1}^{2}+\frac{1}{10}$
$\Rightarrow \Sigma \alpha_{1}^{3}=\frac{1}{10} \Sigma \alpha_{1}^{2}+\Sigma \frac{1}{10}$
$\Rightarrow \Sigma \alpha_{1}^{3}-\frac{1}{10} \Sigma \alpha_{1}^{2}=\frac{3}{10}$
$\therefore \left(\alpha_{1}^{3}+\beta_{1}^{3}+\gamma_{1}^{3}\right)-\frac{1}{10}\left(\alpha_{1}^{2}+\beta_{1}^{2}+\gamma_{1}^{2}\right)=\frac{3}{10}$