If the coefficients of $x^4$ and  $x^2$  in the expansion of

 ${\left(x+\sqrt{x^2-1}\right)}^6+{\left(x-\sqrt{x^2-1}\right)}^6$ are $\alpha$ and $\beta$  respectively, then  $\alpha -\beta$

is equal to 

${\left(x+\sqrt{x^2-1}\right)}^6+{\left(x-\sqrt{x^2-1}\right)}^6$

$\begin{array}{l}=2\left\{{ }^6{C}_0{x}^6{+}^6{C}_2{x}^4\left(x^2-1\right){+}^6{C}_4{x}^2{\left(x^2-1\right)}^2{+}^6{C}_6{\left(x^2-1\right)}^3\right\}\\ =2\left\{x^6+15\left(x^6-x^4\right)+15x^2\left(x^4-2x^2+1\right)+\left(x^6-3x^4+3x^2-1\right)\right\}\\ =2\left(32x^5-48x^4+18x^2-1\right)\end{array}$

$\therefore \alpha =-96$ and $\beta =36$ 

Hence, $\alpha -\beta =-96-36=-132$