Question

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

Answer 1

Answer: 3.00

We have $5x^3-qx-1=0$

Now, $\frac{{\alpha }^2-3}{\beta \gamma }+\frac{{\beta }^2-3}{\gamma \alpha }+\frac{{\gamma }^2-3}{\alpha \beta }$
$=\frac{{\alpha }^3-3\alpha +{\beta }^3-3\beta +{\gamma }^3-3\gamma }{\alpha \beta \gamma }$
$=\frac{{\alpha }^3+{\beta }^3+{\gamma }^3-3(\alpha +\beta +\gamma )}{\alpha \beta \gamma }$
$\left(\because \alpha +\beta +\gamma =0\Rightarrow {\alpha }^3+{\beta }^3+{\gamma }^3=3\alpha \beta \gamma \right)$
$=\frac{3\alpha \beta \gamma }{\alpha \beta \gamma }=3.$
Aliter : $\alpha \beta \gamma =\frac{1}{5}$ and $\alpha +\beta +\gamma =0$
$\therefore \left(\frac{{\alpha }^2-3}{\beta \gamma }+\frac{{\beta }^2-3}{\gamma \alpha }+\frac{{\gamma }^2-3}{\alpha \beta }\right)$
$=5\alpha \left({\alpha }^2-3\right)+5\beta \left({\beta }^2-3\right)+5\gamma \left({\gamma }^2-3\right)$
$=5\left({\alpha }^3+{\beta }^3+{\gamma }^3\right)-15(\alpha +\beta +\gamma )$
$=5(3\alpha \beta \gamma )=15\times \frac{1}{5}=3$
[As $\alpha +\beta +\gamma =0$ ]