If (x+frac{1}{x})^{2}=3,then the value of (x^ | vedclass

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(A) Given $(x+\frac{1}{x})^{2}=3$.Taking the square root on both sides,we get $x+\frac{1}{x}=\sqrt{3}$.Cubing both sides,we get $(x+\frac{1}{x})^{3}=(

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(A) Given $(x+\frac{1}{x})^{2}=3$.Taking the square root on both sides,we get $x+\frac{1}{x}=\sqrt{3}$.Cubing both sides,we get $(x+\frac{1}{x})^{3}=(\sqrt{3})^{3}$.Using the identity $(a+b)^{3}=a^{3}+b^{3}+3ab(a+b)$,we have $x^{3}+\frac{1}{x^{3}}+3(x+\frac{1}{x})=3\sqrt{3}$.Substituting $x+\frac{1}{x}=\sqrt{3}$,we get $x^{3}+\frac{1}{x^{3}}+3\sqrt{3}=3\sqrt{3}$,which implies $x^{3}+\frac{1}{x^{3}}=0$.Multiplying by $x^{3}$,we get $x^{6}+1=0$,so $x^{6}=-1$.Now,substitute $x^{6}=-1$ into the expression $(x^{72}+x^{66}+x^{54}+x^{36}+x^{24}+x^{6}+1)$:$= (x^{6})^{12} + (x^{6})^{11} + (x^{6})^{9} + (x^{6})^{6} + (x^{6})^{4} + x^{6} + 1$$= (-1)^{12} + (-1)^{11} + (-1)^{9} + (-1)^{6} + (-1)^{4} + (-1) + 1$$= 1 - 1 - 1 + 1 + 1 - 1 + 1 = 1$.

If $(x+\frac{1}{x})^{2}=3$,then the value of $(x^{72}+x^{66}+x^{54}+x^{36}+x^{24}+x^{6}+1)$ is

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