Question

If $\alpha, \beta, \gamma$ are the roots of $x^{3}+4 x+1=0$, then the equation whose roots are $\frac{\alpha^{2}}{\beta+\gamma}, \frac{\beta^{2}}{\gamma+\alpha},\,\frac{\gamma^{2}}{\alpha+\beta}$ is

• A. $x^{3}-4 x-1=0$
• B. $x^{3}-4 x+1=0$
• C. $x^{3}+4 x-1=0$
• D. $x^{3}+4 x+1=0$

Answer 1

Answer: (C)

Given, $\alpha, \beta$ and $\gamma$ are the roots of
$x^{3}+4 x+1=0$
$\alpha+\beta+\gamma=0, \alpha \beta+\beta \gamma+\gamma \alpha=4, \alpha \beta \gamma=-1$
Now, $\frac{\alpha^{2}}{\beta+\gamma}+\frac{\beta^{2}}{\gamma+\alpha}+\frac{\gamma^{2}}{\alpha+\beta}=\frac{\alpha^{2}}{-\alpha}+\frac{\beta^{2}}{-\beta}+\frac{\gamma^{2}}{-\gamma}$
$=-(\alpha+\beta+\gamma)=0$
$\frac{\alpha^{2} \beta^{2}}{(\beta+\gamma)(\gamma+\alpha)}+\frac{\beta^{2} \gamma^{2}}{(\gamma+\alpha)(\alpha+\beta)}+\frac{\gamma^{2} \alpha^{2}}{(\beta+\gamma)(\alpha+\beta)}$
$=\alpha \beta+\beta \gamma+\gamma \alpha =4$
and $\frac{\alpha^{2} \beta^{2} \gamma^{2}}{(\beta+\gamma)(\gamma+\alpha)(\alpha+\beta)}=-\alpha \beta \gamma=1$
$(\because \alpha+\beta+\gamma=0)$
$\therefore $ Required equation is
$x^{3}+ 4 x-1=0$