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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
if  x^3  = 1  and  x ≠ 1  simplificar  (((1/x^4 )/(1 + x^5 )))^3
ifx3=1andx1simplificar(1x41+x5)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
x3=1x4=xx5=x2=x3x=1x(1x41+x5)3=(1x(1+1x))3=1(1+x)3=11+3x+3x2+x3=13(x2+x+1)1=13×x31x11=13×01=1
Commented by HongKing last updated on 11/Jan/22
thank you so much my dear Sir cool
thankyousomuchmydearSircool
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
Anotherwayx3=1andx1;(1x41+x5)3=?x31=0(x1)(x2+x+1)=0x2+x+1=0[x1]x3=1x2+x+1=0x3=1:(1x41+x5)3=(1x3x1+x3x2)3=(1x1+x2)3=(1x1x1)3[x2=x1]=(1x2)3=1(x3)2=1

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