If x+frac{1}{x}=3, then the value of x^{6}+fr | vedclass

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

Vedclass · Accepted Answer

$322$

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(D) Given that $x+\frac{1}{x}=3.$Squaring both sides,we get $\left(x+\frac{1}{x}\right)^{2}=3^{2}.$$x^{2}+\frac{1}{x^{2}}+2=9 \Rightarrow x^{2}+\frac{1}{x^{2}}=7.$Now,cubing both sides of the equation $x^{2}+\frac{1}{x^{2}}=7,$ we get $\left(x^{2}+\frac{1}{x^{2}}\right)^{3}=7^{3}.$Using the identity $(a+b)^{3}=a^{3}+b^{3}+3ab(a+b),$$x^{6}+\frac{1}{x^{6}}+3\left(x^{2}+\frac{1}{x^{2}}\right)=343.$Substituting the value $x^{2}+\frac{1}{x^{2}}=7,$$x^{6}+\frac{1}{x^{6}}+3(7)=343.$$x^{6}+\frac{1}{x^{6}}+21=343.$$x^{6}+\frac{1}{x^{6}}=343-21=322.$

If $x+\frac{1}{x}=3,$ then the value of $x^{6}+\frac{1}{x^{6}}$ is

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