The range of $\text{'}\alpha \text{'}$ for which all the abscissa of points of local extrema of the function $f\left(x\right)=x^3-3\alpha x^2+3({\alpha }^2-1)x+1$ lie in the interval $(-2,4),$ is

$f\text{'}\left(x\right)=3x^2-3\alpha \left(2x\right)+3({\alpha }^2-1)=3x^2-6\alpha x+3({\alpha }^2-1)$

Since, extrema lie in (-2, 4) 

So roots of $f\text{'}\left(x\right)$ lie in (-2, 4) 

$f\text{'}\left(x\right)=3(x-(\alpha -1))(x-(\alpha +1))\Rightarrow -2<\alpha -1and\alpha +1<4\Rightarrow -1<\alpha <3$

Second method 

$f\text{'}(-2)>0\Rightarrow 12+12\alpha +3{\alpha }^2-3>0\Rightarrow 3{\alpha }^2+12\alpha +9>0\Rightarrow (\alpha +1)(\alpha +3)>0$

$also,f\text{'}\left(4\right)>0$