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

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Question Number 26046 by ktomboy1992 last updated on 19/Dec/17
x^3 + (1/x^3 )=18 find the valu of x+(1/x)
$${x}^{\mathrm{3}} +\:\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\mathrm{18}\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{valu}\:\mathrm{of}\:{x}+\frac{\mathrm{1}}{{x}} \\ $$
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
$${Given} \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\mathrm{18} \\ $$$$\left({x}+\frac{\mathrm{1}}{{x}}\right)\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{1}\right)=\mathrm{18} \\ $$$$\left({x}+\frac{\mathrm{1}}{{x}}\right)\left\{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{3}\right\}=\mathrm{18} \\ $$$${y}\left({y}^{\mathrm{2}} −\mathrm{3}\right)=\mathrm{18}\:{where}\:{x}+\frac{\mathrm{1}}{{x}}={y},\:{say} \\ $$$${y}^{\mathrm{3}} −\mathrm{3}{y}−\mathrm{18}=\mathrm{0} \\ $$$${y}^{\mathrm{3}} −\mathrm{3}{y}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} −\mathrm{9}{y}+\mathrm{6}{y}−\mathrm{18}=\mathrm{0} \\ $$$${y}^{\mathrm{2}} \left({y}−\mathrm{3}\right)+\mathrm{3}{y}\left({y}−\mathrm{3}\right)+\mathrm{6}\left({y}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\left({y}−\mathrm{3}\right)\left({y}^{\mathrm{2}} +\mathrm{3}{y}+\mathrm{6}\right)=\mathrm{0} \\ $$$$\left({y}−\mathrm{3}\right)\left({y}^{\mathrm{2}} +\mathrm{3}{y}+\mathrm{6}\right)=\mathrm{0} \\ $$$${y}−\mathrm{3}=\mathrm{0} \\ $$$${y}=\mathrm{3} \\ $$$${x}+\frac{\mathrm{1}}{{x}}=\mathrm{3} \\ $$
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)?
$$\mathrm{Why}\:\mathrm{have}\:\mathrm{you}\:\mathrm{left}\:{y}^{\mathrm{2}} +\mathrm{3}{y}+\mathrm{6}=\mathrm{0}? \\ $$$${y}=\frac{-\mathrm{3}\:\pm\sqrt{\mathrm{9}−\mathrm{24}}}{\mathrm{2}} \\ $$$$\:\:=\frac{-\mathrm{3}\:\pm{i}\sqrt{\mathrm{15}}}{\mathrm{2}} \\ $$$${x}+\frac{\mathrm{1}}{{x}}=\frac{-\mathrm{3}\:\pm{i}\sqrt{\mathrm{15}}}{\mathrm{2}} \\ $$$$\mathrm{Is}\:\mathrm{this}\:\mathrm{not}\:\mathrm{a}\:\mathrm{valid}\:\mathrm{value}\:\mathrm{of}\:{x}+\frac{\mathrm{1}}{{x}}? \\ $$
Commented by $@ty@m last updated on 19/Dec/17
assuming y to be real. otherwise 3 possible solutions of course.
$${assuming}\:{y}\:{to}\:{be}\:{real}. \\ $$$${otherwise}\:\mathrm{3}\:{possible}\:{solutions}\: \\ $$$${of}\:{course}. \\ $$
Commented by Rasheed.Sindhi last updated on 19/Dec/17
ThanX !
$$\mathcal{T}{han}\mathcal{X}\:! \\ $$

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