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Question-156992

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

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