Answer:Comparing the equation $x^2-2 \sqrt{3}x+4=0 \ $

with  $ax^2+bx+c=0 \Rightarrow a=1, b=-2\sqrt{3}, c=4.$

We know, $\alpha +\beta=-b/a, \alpha\beta=c/a \Rightarrow \alpha+\beta=2\sqrt{3}, \alpha\beta=4.$

Thus,

 $(\alpha+\beta)^3=\alpha^3+\beta^3+3\alpha\beta(\alpha+\beta) \Rightarrow\\ \alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)\\ =(2 \sqrt{3})^3-3\times4(2 \sqrt{3})=8 \times3 \sqrt{3}-8 \times3 \sqrt{3}=0.$

Thus the answer is 0.