If left(x+frac{1}{x}right)^{2}=3,then the val | vedclass

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

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(A) Given $\left(x+\frac{1}{x}\right)^{2}=3$.Taking the square root on both sides,we get $x+\frac{1}{x}=\sqrt{3}$.Cubing both sides,we get $\left(x+\frac{1}{x}\right)^{3} = (\sqrt{3})^{3}$.Using the identity $(a+b)^{3} = a^{3}+b^{3}+3ab(a+b)$,we have $x^{3}+\frac{1}{x^{3}}+3\left(x+\frac{1}{x}\right) = 3\sqrt{3}$.Substituting $x+\frac{1}{x}=\sqrt{3}$,we get $x^{3}+\frac{1}{x^{3}}+3(\sqrt{3}) = 3\sqrt{3}$.This simplifies to $x^{3}+\frac{1}{x^{3}} = 0$,which implies $x^{6}+1 = 0$.Now,the given expression is $x^{206}+x^{200}+x^{90}+x^{84}+x^{18}+x^{12}+x^{6}+1$.Grouping the terms,we get $x^{200}(x^{6}+1)+x^{84}(x^{6}+1)+x^{12}(x^{6}+1)+1(x^{6}+1)$.Since $x^{6}+1=0$,the entire expression becomes $x^{200}(0)+x^{84}(0)+x^{12}(0)+1(0) = 0$.

If $\left(x+\frac{1}{x}\right)^{2}=3$,then the value of $x^{206}+x^{200}+x^{90}+x^{84}+x^{18}+x^{12}+x^{6}+1$ is

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