The value of (3+sqrt{8})+frac{1}{3-sqrt{8}}-( | vedclass

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(D) First,rationalize the denominator of the fraction $\frac{1}{3-\sqrt{8}}$:$\frac{1}{3-\sqrt{8}} = \frac{1 	imes (3+\sqrt{8})}{(3-\sqrt{8})(3+\sqrt

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(D) First,rationalize the denominator of the fraction $\frac{1}{3-\sqrt{8}}$:$\frac{1}{3-\sqrt{8}} = \frac{1 	imes (3+\sqrt{8})}{(3-\sqrt{8})(3+\sqrt{8})} = \frac{3+\sqrt{8}}{3^2 - (\sqrt{8})^2} = \frac{3+\sqrt{8}}{9-8} = 3+\sqrt{8}$.Now,substitute this back into the original expression:$(3+\sqrt{8}) + (3+\sqrt{8}) - (6+4\sqrt{2})$.Since $\sqrt{8} = \sqrt{4 	imes 2} = 2\sqrt{2}$,we can rewrite the expression as:$(3+2\sqrt{2}) + (3+2\sqrt{2}) - (6+4\sqrt{2})$.$= 6 + 4\sqrt{2} - 6 - 4\sqrt{2} = 0$.

The value of $(3+\sqrt{8})+\frac{1}{3-\sqrt{8}}-(6+4 \sqrt{2})$ is

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