If $$x-\sqrt{3}-\sqrt{2}=0$$ and $$y-\sqrt{3}+\sqrt{2}=0$$ then the value of $$(x^{3}-20\sqrt{2})-(y^{3}+2\sqrt{2})$$

Given, $$x-\sqrt{\ 3}-\sqrt{\ 2}=0$$

$$=$$>   $$x=\sqrt{\ 3}+\sqrt{\ 2}$$

and $$y-\sqrt{\ 3}+\sqrt{\ 2}=0$$

$$=$$>  $$y=\sqrt{\ 3}-\sqrt{\ 2}$$

$$\left(x^3-20\sqrt{\ 2}\right)-\left(y^3+2\sqrt{\ 2}\right)$$ = $$x^3-y^3-20\sqrt{\ 2}-2\sqrt{\ 2}$$                

  = $$\left(\sqrt{\ 3}+\sqrt{\ 2}\right)^3-\left(\sqrt{\ 3}-\sqrt{\ 2}\right)^3-22\sqrt{\ 2}$$

  =  $$3\sqrt{\ 3}+2\sqrt{\ 2}+3.\sqrt{\ 3}.\sqrt{\ 2}\left(\sqrt{\ 3}+\sqrt{\ 2}\right)-3\sqrt{\ 3}+2\sqrt{\ 2}+3.\sqrt{\ 3}.\sqrt{\ 2}\left(\sqrt{\ 3}-\sqrt{\ 2}\right)-22\sqrt{\ 2}$$

  =  $$2\sqrt{\ 2}+9\sqrt{\ 2}+6\sqrt{\ 3}+2\sqrt{\ 2}+9\sqrt{\ 2}-6\sqrt{\ 2}-22\sqrt{\ 2}$$

  = 0