Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 156992 by MathSh last updated on 18/Oct/21

Commented by MJS_new last updated on 19/Oct/21

I get x=1∨x≈−.428605913602 no other solution ∈C

$$\mathrm{I}\:\mathrm{get} \\ $$$${x}=\mathrm{1}\vee{x}\approx−.\mathrm{428605913602} \\ $$$$\mathrm{no}\:\mathrm{other}\:\mathrm{solution}\:\in\mathbb{C} \\ $$

Answered by Rasheed.Sindhi last updated on 18/Oct/21

(3/( ((x+1))^(1/3) ))+(x/( ((x^3 +1))^(1/3) ))=2(4)^(1/3) ((3/( ((x+1))^(1/3) ))+(x/( ((x^3 +1))^(1/3) ))=2(4)^(1/3) )^3 ▶((27)/(x+1))+(x^3 /(x^3 +1)) +3((3/( ((x+1))^(1/3) )))((x/( ((x^3 +1))^(1/3) )))((3/( ((x+1))^(1/3) ))+(x/( ((x^3 +1))^(1/3) )))=8(4) ▶((27)/(x+1))+(x^3 /(x^3 +1))+3((3/( ((x+1))^(1/3) )))((x/( ((x^3 +1))^(1/3) )))(2(4)^(1/3) )=8(4) ▶((27)/(x+1))+(x^3 /(x^3 +1))+(((9x)/( (((x^3 +1)(x+1)))^(1/3) )))(2(4)^(1/3) )=8(4) ▶((18((4 ))^(1/3) x)/( (((x^3 +1)(x+1)))^(1/3) ))=32−((27)/(x+1))−(x^3 /(x^3 +1)) =((32(x+1)(x^2 −x+1)−27(x^2 −x+1)+x^3 )/((x+1)(x^2 −x+1))) =((32(x^3 +1)−27(x^2 −x+1)+x^3 )/((x+1)(x^2 −x+1))) =((33x^3 −27x^2 +27x+5)/(x^3 +1)) ▶(((18((4 ))^(1/3) x)/( (((x^3 +1)(x+1)))^(1/3) )))^3 =(((33x^3 −27x^2 +27x+5)/(x^3 +1)))^3 ((5832(4)x^3 )/((x^3 +1)(x+1)))=(((33x^3 −27x^2 +27x+5)^3 )/((x^3 +1)^3 )) ((5832(4)x^3 )/(x+1))=(((33x^3 −27x^2 +27x+5)^3 )/((x^3 +1)^2 )) ((5832(4)x^3 )/(x+1))=(((33x^3 −27x^2 +27x+5)^3 )/((x+1)^2 (x^2 −x+1)^2 )) 5832(4)x^3 =(((33x^3 −27x^2 +27x+5)^3 )/((x+1)(x^2 −x+1)^2 )) ▶5832(4)x^3 (x+1)(x^2 −x+1)^2 −(33x^3 −27x^2 +27x+5)^3 =0 x=−0.423807 (With the help of calculator)

$$\frac{\mathrm{3}}{\:\sqrt[{\mathrm{3}}]{\mathrm{x}+\mathrm{1}}}+\frac{\mathrm{x}}{\:\sqrt[{\mathrm{3}}]{\mathrm{x}^{\mathrm{3}} +\mathrm{1}}}=\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{4}}\: \\ $$$$\left(\frac{\mathrm{3}}{\:\sqrt[{\mathrm{3}}]{\mathrm{x}+\mathrm{1}}}+\frac{\mathrm{x}}{\:\sqrt[{\mathrm{3}}]{\mathrm{x}^{\mathrm{3}} +\mathrm{1}}}=\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{4}}\:\right)^{\mathrm{3}} \\ $$$$\blacktriangleright\frac{\mathrm{27}}{\mathrm{x}+\mathrm{1}}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{x}^{\mathrm{3}} +\mathrm{1}} \\ $$$$\:\:\:\:\:+\mathrm{3}\left(\frac{\mathrm{3}}{\:\sqrt[{\mathrm{3}}]{\mathrm{x}+\mathrm{1}}}\right)\left(\frac{\mathrm{x}}{\:\sqrt[{\mathrm{3}}]{\mathrm{x}^{\mathrm{3}} +\mathrm{1}}}\right)\left(\frac{\mathrm{3}}{\:\sqrt[{\mathrm{3}}]{\mathrm{x}+\mathrm{1}}}+\frac{\mathrm{x}}{\:\sqrt[{\mathrm{3}}]{\mathrm{x}^{\mathrm{3}} +\mathrm{1}}}\right)=\mathrm{8}\left(\mathrm{4}\right) \\ $$$$\blacktriangleright\frac{\mathrm{27}}{\mathrm{x}+\mathrm{1}}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{x}^{\mathrm{3}} +\mathrm{1}}+\mathrm{3}\left(\frac{\mathrm{3}}{\:\sqrt[{\mathrm{3}}]{\mathrm{x}+\mathrm{1}}}\right)\left(\frac{\mathrm{x}}{\:\sqrt[{\mathrm{3}}]{\mathrm{x}^{\mathrm{3}} +\mathrm{1}}}\right)\left(\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{4}}\right)=\mathrm{8}\left(\mathrm{4}\right) \\ $$$$\blacktriangleright\frac{\mathrm{27}}{\mathrm{x}+\mathrm{1}}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{x}^{\mathrm{3}} +\mathrm{1}}+\left(\frac{\mathrm{9x}}{\:\sqrt[{\mathrm{3}}]{\left(\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)\left(\mathrm{x}+\mathrm{1}\right)}}\right)\left(\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{4}}\right)=\mathrm{8}\left(\mathrm{4}\right) \\ $$$$\blacktriangleright\frac{\mathrm{18}\sqrt[{\mathrm{3}}]{\mathrm{4}\:}\:\mathrm{x}}{\:\sqrt[{\mathrm{3}}]{\left(\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)\left(\mathrm{x}+\mathrm{1}\right)}}=\mathrm{32}−\frac{\mathrm{27}}{\mathrm{x}+\mathrm{1}}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{x}^{\mathrm{3}} +\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{32}\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)−\mathrm{27}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)+\mathrm{x}^{\mathrm{3}} }{\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{32}\left(\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)−\mathrm{27}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)+\mathrm{x}^{\mathrm{3}} }{\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{33x}^{\mathrm{3}} −\mathrm{27x}^{\mathrm{2}} +\mathrm{27x}+\mathrm{5}}{\mathrm{x}^{\mathrm{3}} +\mathrm{1}} \\ $$$$\blacktriangleright\left(\frac{\mathrm{18}\sqrt[{\mathrm{3}}]{\mathrm{4}\:}\:\mathrm{x}}{\:\sqrt[{\mathrm{3}}]{\left(\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)\left(\mathrm{x}+\mathrm{1}\right)}}\right)^{\mathrm{3}} =\left(\frac{\mathrm{33x}^{\mathrm{3}} −\mathrm{27x}^{\mathrm{2}} +\mathrm{27x}+\mathrm{5}}{\mathrm{x}^{\mathrm{3}} +\mathrm{1}}\right)^{\mathrm{3}} \\ $$$$\:\:\:\:\frac{\mathrm{5832}\left(\mathrm{4}\right)\mathrm{x}^{\mathrm{3}} }{\left(\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)\left(\mathrm{x}+\mathrm{1}\right)}=\frac{\left(\mathrm{33x}^{\mathrm{3}} −\mathrm{27x}^{\mathrm{2}} +\mathrm{27x}+\mathrm{5}\right)^{\mathrm{3}} }{\left(\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\:\:\:\:\frac{\mathrm{5832}\left(\mathrm{4}\right)\mathrm{x}^{\mathrm{3}} }{\mathrm{x}+\mathrm{1}}=\frac{\left(\mathrm{33x}^{\mathrm{3}} −\mathrm{27x}^{\mathrm{2}} +\mathrm{27x}+\mathrm{5}\right)^{\mathrm{3}} }{\left(\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\frac{\mathrm{5832}\left(\mathrm{4}\right)\mathrm{x}^{\mathrm{3}} }{\mathrm{x}+\mathrm{1}}=\frac{\left(\mathrm{33x}^{\mathrm{3}} −\mathrm{27x}^{\mathrm{2}} +\mathrm{27x}+\mathrm{5}\right)^{\mathrm{3}} }{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\mathrm{5832}\left(\mathrm{4}\right)\mathrm{x}^{\mathrm{3}} =\frac{\left(\mathrm{33x}^{\mathrm{3}} −\mathrm{27x}^{\mathrm{2}} +\mathrm{27x}+\mathrm{5}\right)^{\mathrm{3}} }{\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\blacktriangleright\mathrm{5832}\left(\mathrm{4}\right)\mathrm{x}^{\mathrm{3}} \left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\left(\mathrm{33x}^{\mathrm{3}} −\mathrm{27x}^{\mathrm{2}} +\mathrm{27x}+\mathrm{5}\right)^{\mathrm{3}} =\mathrm{0} \\ $$$$ \\ $$$$\:\:\:\mathrm{x}=−\mathrm{0}.\mathrm{423807}\:\left({With}\:{the}\:{help}\:{of}\:{calculator}\right) \\ $$

Commented by Tawa11 last updated on 18/Oct/21

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Commented by MathSh last updated on 18/Oct/21

dear Ser ans:1

$$\mathrm{dear}\:\mathrm{Ser}\:\mathrm{ans}:\mathrm{1} \\ $$

Commented by mr W last updated on 18/Oct/21

such questions have no generality! you can′t really solve them. they just request that you “see” the solution or not. but when a single value is changed, no solution can be “seen”. 3+(x/( ((x^2 −x+1))^(1/3) ))=2((4(1+x)))^(1/3) 3+(1/( (((1/x)−(1/x^2 )+(1/x^3 )))^(1/3) ))=2((4(1+x)))^(1/3) one guess 3+1=2×2, and one gets x=1 which fulfills the equation. so x=1 is a solution you can see. whether there is an other solution? you can not know and you can not see. when the eqn. is changed to (5/( ((1+x))^(1/3) ))+(x/( ((1+x^3 ))^(1/3) ))=2(4)^(1/3) no solution can be “seen”.

$${such}\:{questions}\:{have}\:{no}\:{generality}! \\ $$$${you}\:{can}'{t}\:{really}\:{solve}\:{them}.\:{they} \\ $$$${just}\:{request}\:{that}\:{you}\:``{see}''\:{the}\:{solution} \\ $$$${or}\:{not}.\:{but}\:{when}\:{a}\:{single}\:{value}\:{is} \\ $$$${changed},\:{no}\:{solution}\:{can}\:{be}\:``{seen}''. \\ $$$$\mathrm{3}+\frac{{x}}{\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} −{x}+\mathrm{1}}}=\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{4}\left(\mathrm{1}+{x}\right)} \\ $$$$\mathrm{3}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}^{\mathrm{3}} }}}=\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{4}\left(\mathrm{1}+{x}\right)} \\ $$$${one}\:{guess}\:\mathrm{3}+\mathrm{1}=\mathrm{2}×\mathrm{2},\:{and}\:{one}\:{gets}\:{x}=\mathrm{1} \\ $$$${which}\:{fulfills}\:{the}\:{equation}.\:{so}\:{x}=\mathrm{1} \\ $$$${is}\:{a}\:{solution}\:{you}\:{can}\:{see}.\:{whether} \\ $$$${there}\:{is}\:{an}\:{other}\:{solution}?\:{you}\:{can}\: \\ $$$${not}\:{know}\:{and}\:{you}\:{can}\:{not}\:{see}. \\ $$$${when}\:{the}\:{eqn}.\:{is}\:{changed}\:{to} \\ $$$$\frac{\mathrm{5}}{\:\sqrt[{\mathrm{3}}]{\mathrm{1}+{x}}}+\frac{{x}}{\:\sqrt[{\mathrm{3}}]{\mathrm{1}+{x}^{\mathrm{3}} }}=\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{4}} \\ $$$${no}\:{solution}\:{can}\:{be}\:``{seen}''. \\ $$

Commented by mr W last updated on 18/Oct/21

MathSh sir: x=1 is not the only solution!

$${MathSh}\:{sir}: \\ $$$${x}=\mathrm{1}\:{is}\:{not}\:{the}\:{only}\:{solution}! \\ $$

Commented by MathSh last updated on 18/Oct/21

Thank you dear Ser

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{Ser} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com