if-x-3-1-and-x-1-simplificar-1-x-4-1-x-5-3-

Question Number 163897 by HongKing last updated on 11/Jan/22

$$\mathrm{if}\:\:\mathrm{x}^{\mathrm{3}} \:=\:\mathrm{1}\:\:\mathrm{and}\:\:\mathrm{x}\:\neq\:\mathrm{1} \\ $$$$\mathrm{simplificar}\:\:\left(\frac{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }}{\mathrm{1}\:+\:\mathrm{x}^{\mathrm{5}} }\right)^{\mathrm{3}} \\ $$

Answered by mr W last updated on 11/Jan/22

x^3 =1 x^4 =x x^5 =x^2 =(x^3 /x)=(1/x) (((1/x^4 )/(1 + x^5 )))^3 =((1/(x(1+(1/x)))))^3 =(1/((1+x)^3 )) =(1/(1+3x+3x^2 +x^3 ))=(1/(3(x^2 +x+1)−1)) =(1/(3×((x^3 −1)/(x−1))−1))=(1/(3×0−1))=−1

$${x}^{\mathrm{3}} =\mathrm{1} \\ $$$${x}^{\mathrm{4}} ={x} \\ $$$${x}^{\mathrm{5}} ={x}^{\mathrm{2}} =\frac{{x}^{\mathrm{3}} }{{x}}=\frac{\mathrm{1}}{{x}} \\ $$$$\:\:\left(\frac{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }}{\mathrm{1}\:+\:\mathrm{x}^{\mathrm{5}} }\right)^{\mathrm{3}} =\left(\frac{\mathrm{1}}{{x}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)}\right)^{\mathrm{3}} =\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{3}{x}+\mathrm{3}{x}^{\mathrm{2}} +{x}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{3}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)−\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}×\frac{{x}^{\mathrm{3}} −\mathrm{1}}{{x}−\mathrm{1}}−\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{3}×\mathrm{0}−\mathrm{1}}=−\mathrm{1} \\ $$

Commented by HongKing last updated on 11/Jan/22

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{cool} \\ $$

Answered by Rasheed.Sindhi last updated on 14/Jan/22

An other way x^3 = 1 and x ≠ 1;(((1/x^4 )/(1 + x^5 )))^3 =? x^3 −1=0⇒(x−1)(x^2 +x+1)=0 ⇒x^2 +x+1=0 [∵ x≠1] determinant (((x^3 = 1 ∧ x^2 +x+1=0))) x^3 = 1: (((1/x^4 )/(1 + x^5 )))^3 =(((1/(x^3 ∙x))/(1 + x^3 ∙x^2 )))^3 =(((1/x)/(1 + x^2 )))^3 =(((1/x)/(1−x−1)))^3 [∵ x^2 =−x−1] =(−(1/x^2 ))^3 =−(1/((x^3 )^2 ))=−1

$$\mathrm{An}\:\mathrm{other}\:\mathrm{way} \\ $$$$\mathrm{x}^{\mathrm{3}} \:=\:\mathrm{1}\:\:\mathrm{and}\:\:\mathrm{x}\:\neq\:\mathrm{1};\left(\frac{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }}{\mathrm{1}\:+\:\mathrm{x}^{\mathrm{5}} }\right)^{\mathrm{3}} =?\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\mathrm{x}^{\mathrm{3}} −\mathrm{1}=\mathrm{0}\Rightarrow\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\Rightarrow\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}=\mathrm{0}\:\:\:\left[\because\:\mathrm{x}\neq\mathrm{1}\right] \\ $$$$\begin{array}{|c|}{\mathrm{x}^{\mathrm{3}} \:=\:\mathrm{1}\:\:\:\:\wedge\:\:\:\:\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}=\mathrm{0}}\\\hline\end{array} \\ $$$$\mathrm{x}^{\mathrm{3}} \:=\:\mathrm{1}:\:\left(\frac{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }}{\mathrm{1}\:+\:\mathrm{x}^{\mathrm{5}} }\right)^{\mathrm{3}} =\left(\frac{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} \centerdot\mathrm{x}}}{\mathrm{1}\:+\:\mathrm{x}^{\mathrm{3}} \centerdot\mathrm{x}^{\mathrm{2}} }\right)^{\mathrm{3}} =\left(\frac{\frac{\mathrm{1}}{\mathrm{x}}}{\mathrm{1}\:+\:\mathrm{x}^{\mathrm{2}} }\right)^{\mathrm{3}} \\ $$$$=\left(\frac{\frac{\mathrm{1}}{\mathrm{x}}}{\cancel{\mathrm{1}}−\mathrm{x}−\cancel{\mathrm{1}}}\right)^{\mathrm{3}} \:\:\:\:\:\:\:\:\:\left[\because\:\mathrm{x}^{\mathrm{2}} =−\mathrm{x}−\mathrm{1}\right] \\ $$$$=\left(−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)^{\mathrm{3}} =−\frac{\mathrm{1}}{\left(\mathrm{x}^{\mathrm{3}} \right)^{\mathrm{2}} }=−\mathrm{1} \\ $$

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