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If $x-\frac{1}{x}=2$ , then $(x-\frac{1}{x})^3=8$ .

$(x-\frac{1}{x})^3=x^3-\frac{1}{x^3}-3(x-\frac{1}{x})=x^3-\frac{1}{x^3}-6=8$ .

$\implies$ $x^3-\frac{1}{x^3}=\boxed{\large{14}}$ .

It was given, x - $\frac{1}{x}$ = 2

x³ - $\frac{1}{x³}$

= (x)² - ( $\frac{1}{x}$ )³

= (x - $\frac{1}{x}$ )³ + 3 . x . $\frac{1}{x}$ (x - $\frac{1}{x}$ )

= (2)³ + 3 × 2

= 8 + 6

= 14 (Answer)