If $$x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$$ then $$x^{3} + \frac{1}{x^{3}}$$ is equal to

$$x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$$

=> $$x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} * \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$$

=> $$x= 5+2\sqrt{6}$$ ---------------Eqn(1)

Now, $$\frac{1}{x}=\frac{1}{5+2\sqrt{6}}$$

=> $$\frac{1}{x} = \frac{1}{5+2\sqrt{6}} * \frac{5-2\sqrt{6}}{5-2\sqrt{6}}$$

=> $$\frac{1}{x}= 5-2\sqrt{6}$$ ---------------Eqn(2)

Now, cubing eqns (1)&(2), we get :

=> $$x^3 = 125+72\sqrt{6}+150\sqrt{6}+360 = 485+222\sqrt{6}$$

and $$\frac{1}{x^3} = 125-72\sqrt{6}-150\sqrt{6}+360 = 485-222\sqrt{6}$$

To find : $$x^{3} + \frac{1}{x^{3}}$$

= $$485+222\sqrt{6} + 485-222\sqrt{6}$$

= 970