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x-3-1-x-3-18-find-the-valu-of-x-1-x-




Question Number 26046 by ktomboy1992 last updated on 19/Dec/17
x^3 + (1/x^3 )=18  find the valu of x+(1/x)
x3+1x3=18findthevaluofx+1x
Answered by $@ty@m last updated on 19/Dec/17
Given  x^3 +(1/x^3 )=18  (x+(1/x))(x^2 +(1/x^2 )−1)=18  (x+(1/x)){(x+(1/x))^2 −3}=18  y(y^2 −3)=18 where x+(1/x)=y, say  y^3 −3y−18=0  y^3 −3y^2 +3y^2 −9y+6y−18=0  y^2 (y−3)+3y(y−3)+6(y−3)=0  (y−3)(y^2 +3y+6)=0  (y−3)(y^2 +3y+6)=0  y−3=0  y=3  x+(1/x)=3
Givenx3+1x3=18(x+1x)(x2+1x21)=18(x+1x){(x+1x)23}=18y(y23)=18wherex+1x=y,sayy33y18=0y33y2+3y29y+6y18=0y2(y3)+3y(y3)+6(y3)=0(y3)(y2+3y+6)=0(y3)(y2+3y+6)=0y3=0y=3x+1x=3
Commented by Rasheed.Sindhi last updated on 19/Dec/17
Why have you left y^2 +3y+6=0?  y=((-3 ±(√(9−24)))/2)    =((-3 ±i(√(15)))/2)  x+(1/x)=((-3 ±i(√(15)))/2)  Is this not a valid value of x+(1/x)?
Whyhaveyoulefty2+3y+6=0?y=3±9242=3±i152x+1x=3±i152Isthisnotavalidvalueofx+1x?
Commented by $@ty@m last updated on 19/Dec/17
assuming y to be real.  otherwise 3 possible solutions   of course.
assumingytobereal.otherwise3possiblesolutionsofcourse.
Commented by Rasheed.Sindhi last updated on 19/Dec/17
ThanX !
ThanX!

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