1x 1x 2+1x 2x 323

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$\\Solution :\\Given:\\\\ x+\frac{1}{x}=3\\\\ Now,\\\\ (x+\frac1x)^2=x^2+\frac{1}{x^2}+2x(\frac 1x)\ \ \ \ \ \ \ [\because (a+b)^2=a^2+b^2+2ab]\\\\\\ \Rightarrow 3^2 =x^2+\frac{1}{x^2}+2\\\\\\ \Rightarrow x^2+\frac{1}{x^2}=9-2\\\\\\ \therefore x^2+\frac{1}{x^2}=7\\\\\\ Now,\\ (x^2+\frac{1}{x^2})^3=(x^2)^3+(\frac{1}{x^2})^3+3x^2(\frac 1{x^2})(x^2+\frac 1{x^2})\ \ \ \ \ [\because (a+b)^3=a^3+b^3+3ab(a+b)]\\\\\\ \Rightarrow 7^3=x^6+\frac{1}{x^6}+3\times7\\\\\\ \Rightarrow 343=x^6+\frac{1}{x^6}+21\\\\\\ \Rightarrow x^6+\frac{1}{x^6}=343-21\\\\\\ \therefore x^6+\frac{1}{x^6}=322$

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1x 1x 2+1x 2x 323

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