Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

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 !

$$\mathcal{T}{han}\mathcal{X}\:! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com