If x+frac{1}{x}=3, then the value of x^{8}+fr | vedclass

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(D) Given $x+\frac{1}{x}=3.$Squaring both sides,we get $\left(x+\frac{1}{x}\right)^{2}=3^{2} \Rightarrow x^{2}+\frac{1}{x^{2}}+2=9 \Rightarrow x^{2}+\

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(D) Given $x+\frac{1}{x}=3.$Squaring both sides,we get $\left(x+\frac{1}{x}\right)^{2}=3^{2} \Rightarrow x^{2}+\frac{1}{x^{2}}+2=9 \Rightarrow x^{2}+\frac{1}{x^{2}}=7.$Now,squaring again,$\left(x^{2}+\frac{1}{x^{2}}\right)^{2}=7^{2} \Rightarrow x^{4}+\frac{1}{x^{4}}+2=49 \Rightarrow x^{4}+\frac{1}{x^{4}}=47.$Finally,squaring once more,$\left(x^{4}+\frac{1}{x^{4}}\right)^{2}=47^{2} \Rightarrow x^{8}+\frac{1}{x^{8}}+2=2209 \Rightarrow x^{8}+\frac{1}{x^{8}}=2207.$

If $x+\frac{1}{x}=3,$ then the value of $x^{8}+\frac{1}{x^{8}}$ will be

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