pramrok62

2022-09-03

$\alpha ,\beta ,\gamma$ are roots of cubic equation ${x}^{3}+4x-1=0$

Samantha Braun

$\frac{\left(\beta +1\right)\left(\gamma +1\right)}{{\alpha }^{2}}+\frac{\left(\gamma +1\right)\left(\alpha +1\right)}{{\beta }^{2}}+\frac{\left(\alpha +1\right)\left(\beta +1\right)}{{\gamma }^{2}}$
$=6\left\{\frac{1}{\left(\alpha +1\right){\alpha }^{2}}+\frac{1}{\left(\beta +1\right){\beta }^{2}}+\frac{1}{\left(\gamma +1\right){\gamma }^{2}}\right\}.$
Here, by Vieta's formulas, notice that
$\frac{7}{6}=\frac{1}{\alpha +1}+\frac{1}{\beta +1}+\frac{1}{\gamma +1}=\frac{\left(\alpha \beta +\beta \gamma +\gamma \alpha \right)+2\left(\alpha +\beta +\gamma \right)+3}{\left(\alpha +1\right)\left(\beta +1\right)\left(\gamma +1\right)}.$
Since $\alpha +\beta +\gamma =0,$ we have
$\left(\alpha +1\right)\left(\beta +1\right)\left(\gamma +1\right)=6.$
Noting that your last $\left\{\right\}$ equals to
$\frac{\left({\alpha }^{3}+{\beta }^{3}+{\gamma }^{3}\right)+\left({\alpha }^{2}+{\beta }^{2}+{\gamma }^{2}\right)}{\left(\alpha \beta \gamma {\right)}^{2}\left(\alpha +1\right)\left(\beta +1\right)\left(\gamma +1\right)},$
we can use
${\alpha }^{3}+{\beta }^{3}+{\gamma }^{3}=3\alpha \beta \gamma +\left(\alpha +\beta +\gamma \right)\left({\alpha }^{2}+{\beta }^{2}+{\gamma }^{2}-\alpha \beta -\beta \gamma -\gamma \alpha \right).$
with
$\alpha +\beta +\gamma =0,\alpha \beta +\beta \gamma +\gamma \alpha =4,\alpha \beta \gamma =1$
and
${\alpha }^{2}+{\beta }^{2}+{\gamma }^{2}=\left(\alpha +\beta +\gamma {\right)}^{2}-2\left(\alpha \beta +\beta \gamma +\gamma \alpha \right).$
Now you'll be able to get the answer.

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