We know that ${\left(a+b\right)}^n-{\left(a-b\right)}^n=2\left[{}^{n}C_1{a}^{n-1}b+{}^{n}C_3{a}^{n-3}{b}^3+{}^{n}C_5{a}^{n-5}{b}^5+...\right]$

Given $(\sqrt{3}+i{)}^8-(\sqrt{3}-i{)}^8=\alpha +i\beta$

$\Rightarrow 2\left[{}^{8}C_1{\left(\sqrt{3}\right)}^7\left(i\right)+{}^{8}C_3{\left(\sqrt{3}\right)}^5{\left(i\right)}^3+{}^{8}C_5{\left(\sqrt{3}\right)}^3{\left(i\right)}^5+{}^{8}C_7\left(\sqrt{3}\right){\left(i\right)}^7\right]=\alpha +i\beta$

Using ${}^{n}C_r=\frac{n!}{r!\times \left(n-r\right)!}, i^{4n}=1, i^{4n+1}=i, i^{4n+2}=-1$ and $i^{4n+3}=-i,$

$\Rightarrow 2\left[\frac{8!}{1!\times 7!}\left(27\sqrt{3}\right)\left(i\right)+\frac{8!}{3!\times 5!}\left(9\sqrt{3}\right)\left(-i\right)+\frac{8!}{5!\times 3!}\left(3\sqrt{3}\right)\left(i\right)+\frac{8!}{7!\times 1!}\left(\sqrt{3}\right)\left(-i\right)\right]=\alpha +i\beta$

$\Rightarrow 2\left[8\times \left(27\sqrt{3}\right)\left(i\right)+56\times \left(9\sqrt{3}\right)\left(-i\right)+56\times \left(3\sqrt{3}\right)\left(i\right)+8\times \left(\sqrt{3}\right)\left(-i\right)\right]=\alpha +i\beta$

$\Rightarrow 2\left[216\sqrt{3}i-504\sqrt{3}i+168\sqrt{3}i-8\sqrt{3}i\right]=\alpha +i\beta$

$\Rightarrow -256\sqrt{3}i=\alpha +i\beta$

$\Rightarrow \alpha =0, \beta =-256\sqrt{3}.$

Hence, $\alpha -\frac{\sqrt{3}}{2}\beta =0-\frac{\sqrt{3}}{2}\times \left(-256\sqrt{3}\right)=384.$