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 ${\left(-\frac{1}{x}\right)}^3+\left(-\frac{1}{x}\right)+10=0$
$\Rightarrow 10x^3-x^2-1=0$
$\Rightarrow 10{\alpha }_1^3-{\alpha }_1^2-1=0$
$\Rightarrow {\alpha }_1^3=\frac{1}{10}{\alpha }_1^2+\frac{1}{10}$
$\Rightarrow \Sigma {\alpha }_1^3=\frac{1}{10}\Sigma {\alpha }_1^2+\Sigma \frac{1}{10}$
$\Rightarrow \Sigma {\alpha }_1^3-\frac{1}{10}\Sigma {\alpha }_1^2=\frac{3}{10}$
$\therefore \left({\alpha }_1^3+{\beta }_1^3+{\gamma }_1^3\right)-\frac{1}{10}\left({\alpha }_1^2+{\beta }_1^2+{\gamma }_1^2\right)=\frac{3}{10}$