Question

If $\alpha, \beta$ and $\gamma$ are the roots of the equation $5 x^{3}-q x-1=0,(q \in R)$ 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

We have $5 x^{3}-q x-1=0$

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 $\alpha+\beta+\gamma=0$ ]