If (x+frac{1}{x})=5, what is the value of (x^ | vedclass

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(B) Given $(x+\frac{1}{x})=5.$First,find $(x^{2}+\frac{1}{x^{2}})$ by squaring both sides:$(x+\frac{1}{x})^{2} = 5^{2}$$x^{2}+\frac{1}{x^{2}}+2 = 25$$

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$2525$

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(B) Given $(x+\frac{1}{x})=5.$First,find $(x^{2}+\frac{1}{x^{2}})$ by squaring both sides:$(x+\frac{1}{x})^{2} = 5^{2}$$x^{2}+\frac{1}{x^{2}}+2 = 25$$x^{2}+\frac{1}{x^{2}} = 23.$Next,find $(x^{3}+\frac{1}{x^{3}})$ by cubing both sides:$(x+\frac{1}{x})^{3} = 5^{3}$$x^{3}+\frac{1}{x^{3}}+3(x+\frac{1}{x}) = 125$$x^{3}+\frac{1}{x^{3}}+3(5) = 125$$x^{3}+\frac{1}{x^{3}} = 125-15 = 110.$Now,multiply the two results:$(x^{2}+\frac{1}{x^{2}})(x^{3}+\frac{1}{x^{3}}) = 23 	imes 110$$x^{5}+\frac{1}{x}+x+\frac{1}{x^{5}} = 2530$$(x^{5}+\frac{1}{x^{5}}) + (x+\frac{1}{x}) = 2530$$(x^{5}+\frac{1}{x^{5}}) + 5 = 2530$$x^{5}+\frac{1}{x^{5}} = 2530-5 = 2525.$

If $(x+\frac{1}{x})=5,$ what is the value of $(x^{5}+\frac{1}{x^{5}})?$

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