If x+frac{1}{x}=5, then x^{6}+frac{1}{x^{6}} | vedclass

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(A) Given $x+\frac{1}{x}=5$.First,find $x^3+\frac{1}{x^3}$ by cubing both sides:$(x+\frac{1}{x})^3 = 5^3$$x^3+\frac{1}{x^3}+3(x)(\frac{1}{x})(x+\frac{

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$12098$

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(A) Given $x+\frac{1}{x}=5$.First,find $x^3+\frac{1}{x^3}$ by cubing both sides:$(x+\frac{1}{x})^3 = 5^3$$x^3+\frac{1}{x^3}+3(x)(\frac{1}{x})(x+\frac{1}{x}) = 125$$x^3+\frac{1}{x^3}+3(5) = 125$$x^3+\frac{1}{x^3} = 125-15 = 110$.Now,square both sides to find $x^6+\frac{1}{x^6}$:$(x^3+\frac{1}{x^3})^2 = 110^2$$x^6+\frac{1}{x^6}+2(x^3)(\frac{1}{x^3}) = 12100$$x^6+\frac{1}{x^6}+2 = 12100$$x^6+\frac{1}{x^6} = 12100-2 = 12098$.

If $x+\frac{1}{x}=5,$ then $x^{6}+\frac{1}{x^{6}}$ is

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