Question

If $\alpha ,\beta ,\gamma$ are the zeroes of the cubic polynomial $x^3+4x+2$, then find the value of
$\frac{1}{\alpha +\beta }+\frac{1}{\beta +\gamma }+\frac{1}{\gamma +\alpha }$

Answer 1

We have,

$P\left(x\right)=x^3+4x+2$

Since, $\alpha ,\beta ,\gamma$ are the zeroes of this polynomial.

Now,

$\alpha +\beta +\gamma =-\frac{0}{1}=0$

$\alpha \beta +\beta \gamma +\gamma \alpha =\frac{4}{1}=4$

$\alpha \beta \gamma =-\frac{2}{1}=-2$

Since,

$\frac{1}{\alpha +\beta }+\frac{1}{\beta +\gamma }+\frac{1}{\gamma +\alpha }$

$\Rightarrow \frac{(\beta +\gamma )(\gamma +\alpha )+(\alpha +\beta )(\gamma +\alpha )+(\alpha +\beta )(\beta +\gamma )}{(\alpha +\beta )(\beta +\gamma )(\gamma +\alpha )}$

$\Rightarrow \frac{\beta \gamma +\alpha \beta +{\gamma }^2+\alpha \gamma +\alpha \gamma +{\alpha }^2+\beta \gamma +\alpha \beta +\alpha \beta +\alpha \gamma +{\beta }^2+\beta \gamma }{(\alpha +\beta )(\beta +\gamma )(\gamma +\alpha )}$

$\Rightarrow \frac{{\alpha }^2+{\beta }^2+{\gamma }^2+3(\alpha \beta +\beta \gamma +\gamma \alpha )}{(\alpha +\beta )(\beta +\gamma )(\gamma +\alpha )}$

$\Rightarrow \frac{{\alpha }^2+{\beta }^2+{\gamma }^2+2(\alpha \beta +\beta \gamma +\gamma \alpha )+(\alpha \beta +\beta \gamma +\gamma \alpha )}{(\alpha +\beta )(\beta +\gamma )(\gamma +\alpha )}$

$\Rightarrow \frac{(\alpha +\beta +\gamma {)}^2+(\alpha \beta +\beta \gamma +\gamma \alpha )}{(\alpha +\beta )(\beta +\gamma )(\gamma +\alpha )}$

$\Rightarrow \frac{0+0}{(\alpha +\beta )(\beta +\gamma )(\gamma +\alpha )}$

$\Rightarrow 0$

Hence, this is the answer.