Question

Suppose $\alpha, \beta, \gamma $ are the roots of
$x^{3}+x^{2}+x+2=0 .$ Then, the value of
$\left(\frac{\alpha+\beta-2 \gamma}{\gamma}\right)\left(\frac{\beta+\gamma-2 \alpha}{\alpha}\right)\left(\frac{\gamma+\alpha-2 \beta}{\beta}\right)$ is

• A. $-\frac{47}{2}$
• B. $\frac{47}{2}$
• C. $-47$
• D. 47

Answer 1

Answer: (A)

Since, $\alpha, \beta, \gamma$ are the roots of $x^{3}+x^{2}+x+2=0$
$\therefore \alpha+\beta+\gamma=-1$
Now, $\frac{\alpha+\beta-2 \gamma}{\gamma}=\frac{-1-\gamma-2 \gamma}{\gamma}=\frac{-1-3 \gamma}{\gamma}$
$\therefore x=\frac{-1-3 \gamma}{\gamma} \Rightarrow \gamma=\frac{-1}{x+3}$
Now, the equation whose root is $\frac{-1}{x+3}$, is
$\left(\frac{-1}{x+3}\right)^{3}+\left(\frac{-1}{x+3}\right)^{2}+\left(\frac{-1}{x+3}\right)+2=0$
$\Rightarrow -1+(x+3)-(x+3)^{2}+2(x+3)^{3}=0$
$\Rightarrow -1+x+3-x^{2}-9-6 x+2 x^{3}+54 +18 x^{2}+54 x=0$
$\Rightarrow 2 x^{3}+17 x^{2}+49 x+47=0$
$\therefore \left(\frac{\alpha+\beta-2 \gamma}{\gamma}\right)\left(\frac{\beta+\gamma-2 \alpha}{\alpha}\right)\left(\frac{\gamma+\alpha-2 \beta}{\beta}\right)=\frac{-47}{2}$