If x-sqrt{3}-sqrt{2}=0 and y-sqrt{3}+sqrt{2}= | vedclass

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(B) Given equations are $x = \sqrt{3} + \sqrt{2}$ and $y = \sqrt{3} - \sqrt{2}$.First,calculate $(x - y)$:$x - y = (\sqrt{3} + \sqrt{2}) - (\sqrt{3} -

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(B) Given equations are $x = \sqrt{3} + \sqrt{2}$ and $y = \sqrt{3} - \sqrt{2}$.First,calculate $(x - y)$:$x - y = (\sqrt{3} + \sqrt{2}) - (\sqrt{3} - \sqrt{2}) = 2\sqrt{2}$.Next,calculate $(xy)$:$xy = (\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = (\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1$.We need to evaluate $(x^3 - 20\sqrt{2}) - (y^3 + 2\sqrt{2}) = x^3 - y^3 - 22\sqrt{2}$.Using the identity $x^3 - y^3 = (x - y)^3 + 3xy(x - y)$:$x^3 - y^3 = (2\sqrt{2})^3 + 3(1)(2\sqrt{2})$$x^3 - y^3 = 16\sqrt{2} + 6\sqrt{2} = 22\sqrt{2}$.Substituting this into the expression:$22\sqrt{2} - 22\sqrt{2} = 0$.

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

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