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Solution
x=$$2-\sqrt{\ 3}$$
Now $$x^3-x^{-3}$$ will be
Now $$\frac{1}{x}=\frac{1}{2-\sqrt{\ 3}}=\frac{\left(2+\sqrt{\ 3}\right)}{4-3}=2+\sqrt{\ 3}$$
Now $$x-\frac{1}{x}=\ -2\sqrt{\ 3}$$
Taking cube on both sides
we get
$$x^3-\frac{1}{x^3}-3\left(x-\frac{1}{x}\right)=\ -24\sqrt{\ 3}$$
we get $$x^3-\frac{1}{x^3}\ =-24\sqrt{\ 3}-3\left(2\sqrt{\ 3}\right)=-30\sqrt{\ 3}$$
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