Question

If $\alpha ,\beta and\gamma$ are the roots of the equation $x^3$ + 3$x^2$ + 5x - 6 = 0, find the value of $(\alpha -\frac{1}{\beta \gamma })(\beta -\frac{1}{\gamma \alpha })(\gamma -\frac{1}{\alpha \beta }){(\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma })}^{-1}$

• A. $\frac{125}{36}$
• B. $\frac{25}{6}$
• C. $\frac{36}{25}$
• D. $\frac{36}{125}$

Answer 1

The correct option is B

$\frac{25}{6}$

$\alpha ,\beta and\gamma$ are the roots of the equation $x^3$ + 3$x^2$ + 5x - 6 = 0

As we know,
$\alpha +\beta +\gamma =\frac{-b}{a}=-3,$
$\alpha \beta +\beta \gamma +\gamma \alpha =5,$
$\alpha \beta \gamma =\frac{-d}{a}=6$

Here,
$(\alpha -\frac{1}{\beta \gamma })(\beta -\frac{1}{\gamma \alpha })(\gamma -\frac{1}{\alpha \beta }){(\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma })}^{-1}$
$=(\alpha -\frac{1}{\beta \gamma })(\beta -\frac{\gamma }{\gamma a})(\gamma -\frac{1}{\alpha \beta })\left(\frac{\alpha \beta \gamma }{\alpha \beta +\beta \gamma +\gamma \alpha }\right)=\frac{(\alpha \beta \gamma -1)(\alpha \beta \gamma -1)(\alpha \beta \gamma -1)}{{\alpha }^2{\beta }^2{\gamma }^2}\times \frac{\alpha \beta \gamma }{(\alpha \beta +\beta \gamma +\gamma \alpha )}$
$=\frac{{(\alpha \beta \gamma -1)}^3}{\alpha \beta \gamma (\alpha \beta +\beta \gamma +\gamma \alpha )}=\frac{5^3}{6\times 5}=\frac{25}{6}$