Putting $y=\left(\alpha -\beta \right)\left(\alpha -\gamma \right)$, we get, 
$y={\alpha }^2-\alpha \left(\beta +\gamma \right)+\beta \gamma$
Now, $\alpha +\beta +\gamma =0$ and $\alpha \beta \gamma =-2$
$\Rightarrow y={\alpha }^2-\alpha \left(-\alpha \right)+\frac{\left(-2\right)}{\alpha }$
$=2{\alpha }^2-\frac{2}{\alpha }=\frac{2\left({\alpha }^3-1\right)}{\alpha }$
Also, $\alpha$ is a root of the given equation.
$\Rightarrow {\alpha }^3=-\alpha -2$
$\Rightarrow y=\frac{2\left(-\alpha -2-1\right)}{\alpha }=2\left(\frac{-\alpha -3}{\alpha }\right)=-2-\frac{6}{\alpha }$
$\Rightarrow y+2=-\frac{6}{\alpha }\Rightarrow \alpha =-\frac{6}{y+2}$
$\alpha$ is a root of the given equation

$\Rightarrow {\left(\frac{-6}{y+2}\right)}^3+\left(\frac{-6}{y+2}\right)+2=0$
$-216-6{\left(y+2\right)}^2+2{\left(y+2\right)}^3=0$
$\Rightarrow {\left(y+2\right)}^3-3{\left(y+2\right)}^2-108=0$
$\Rightarrow y^3+6y^2+12y+8-3y^2-12-12y-108=0$
$\Rightarrow y^3+3y^2-112=0$
Hence, the required equation is $x^3+3x^2-112=0$