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Question Number 68549 by gunawan last updated on 13/Sep/19
y=(x^3 /(x^2 +1)) y^(−1) =...
$${y}=\frac{{x}^{\mathrm{3}} }{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$${y}^{−\mathrm{1}} =… \\ $$
Commented by mathmax by abdo last updated on 13/Sep/19
y=(x^3 /(x^2 +1)) ⇒yx^2 +y =x^3 ⇒x^3 −yx^2 −y =0 (e) y=f(x)⇒x=f^(−1) (y) x =t+(y/3) ⇒x^3 =t^3 +3t^2 (y/3) +3t (y^2 /9)+(y^3 /(27)) =t^3 +yt^2 +(y^2 /3)t +(y^3 /(27)) (e) ⇒t^3 +yt^2 +(y^2 /3)t +(y^3 /(27))−y(t^2 +2t(y/3)+(y^2 /9))−y =0 ⇒ t^3 +yt^2 +(y^2 /3)t +(y^3 /(27))−yt^2 −2(y^2 /3)t −(y^3 /9)−y =0 ⇒ t^3 −(y^2 /3)t −((2y^3 )/(27))−y =0 ⇒t^3 −(y^2 /3)t −((29y^3 )/(27)) =0 (e^′ ) t=u+v (e^′ ) ⇒(u+v)^3 −(y^2 /3)(u+v)−((29y^3 )/(27)) =0 ⇒ u^3 +v^3 +3uv(u+v)−(y^2 /3)(u+v) −((29y^3 )/(27)) =0 ⇒ u^3 +v^3 −((29y^3 )/(27)) +(u+v)(3uv−(y^2 /3)) =0 ⇒u^3 +v^3 =((29y^3 )/(27)) and uv =(y^2 /9) ⇒u^3 .v^3 =(y^6 /9^3 ) so u^3 and v^3 are roots of the equation X^2 −((29y^3 )/(27)) X+(y^6 /9^3 ) =0 Δ =(−((29)/(27))y^3 )^2 −((4y^6 )/9^3 ) =((((29)/(27)))^2 −(4/9^3 ))y^6 X_1 =((((29)/(27))y^3 +(√Δ))/2) and X_2 =((((29)/(27))y^3 −(√Δ))/2) ⇒u =^3 (√X_1 ) and v=^3 (√X_2 ) u=^3 (√X_2 ) and v=^3 (√X_1 ) if we take t =^3 (√X_1 ) +^3 (√X_2 ) ⇒x =f^(−1) (y) =t+(y/3) =^3 (√((((29)/(27))y^3 +(√(((((29)/(27)))^2 −(4/9^3 ))))y^3 )/2)) +^3 (√((((29)/(27))y^3 −(√(((((29)/(27)))^2 −(4/9^3 ))))y^3 )/2)) +(y/3)
$${y}=\frac{{x}^{\mathrm{3}} }{{x}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow{yx}^{\mathrm{2}} \:+{y}\:={x}^{\mathrm{3}} \:\Rightarrow{x}^{\mathrm{3}} −{yx}^{\mathrm{2}} −{y}\:=\mathrm{0}\:\:\left({e}\right)\:{y}={f}\left({x}\right)\Rightarrow{x}={f}^{−\mathrm{1}} \left({y}\right) \\ $$$${x}\:={t}+\frac{{y}}{\mathrm{3}}\:\Rightarrow{x}^{\mathrm{3}} ={t}^{\mathrm{3}} \:+\mathrm{3}{t}^{\mathrm{2}} \frac{{y}}{\mathrm{3}}\:+\mathrm{3}{t}\:\frac{{y}^{\mathrm{2}} }{\mathrm{9}}+\frac{{y}^{\mathrm{3}} }{\mathrm{27}}\:={t}^{\mathrm{3}} \:+{yt}^{\mathrm{2}} \:+\frac{{y}^{\mathrm{2}} }{\mathrm{3}}{t}\:+\frac{{y}^{\mathrm{3}} }{\mathrm{27}} \\ $$$$\left({e}\right)\:\Rightarrow{t}^{\mathrm{3}} \:+{yt}^{\mathrm{2}} \:+\frac{{y}^{\mathrm{2}} }{\mathrm{3}}{t}\:+\frac{{y}^{\mathrm{3}} }{\mathrm{27}}−{y}\left({t}^{\mathrm{2}} +\mathrm{2}{t}\frac{{y}}{\mathrm{3}}+\frac{{y}^{\mathrm{2}} }{\mathrm{9}}\right)−{y}\:=\mathrm{0}\:\Rightarrow \\ $$$${t}^{\mathrm{3}} \:+{yt}^{\mathrm{2}} \:+\frac{{y}^{\mathrm{2}} }{\mathrm{3}}{t}\:+\frac{{y}^{\mathrm{3}} }{\mathrm{27}}−{yt}^{\mathrm{2}} −\mathrm{2}\frac{{y}^{\mathrm{2}} }{\mathrm{3}}{t}\:−\frac{{y}^{\mathrm{3}} }{\mathrm{9}}−{y}\:=\mathrm{0}\:\Rightarrow \\ $$$${t}^{\mathrm{3}} −\frac{{y}^{\mathrm{2}} }{\mathrm{3}}{t}\:\:−\frac{\mathrm{2}{y}^{\mathrm{3}} }{\mathrm{27}}−{y}\:=\mathrm{0}\:\Rightarrow{t}^{\mathrm{3}} −\frac{{y}^{\mathrm{2}} }{\mathrm{3}}{t}\:−\frac{\mathrm{29}{y}^{\mathrm{3}} }{\mathrm{27}}\:=\mathrm{0}\:\:\left({e}^{'} \right) \\ $$$${t}={u}+{v}\:\:\:\left({e}^{'} \right)\:\Rightarrow\left({u}+{v}\right)^{\mathrm{3}} −\frac{{y}^{\mathrm{2}} }{\mathrm{3}}\left({u}+{v}\right)−\frac{\mathrm{29}{y}^{\mathrm{3}} }{\mathrm{27}}\:=\mathrm{0}\:\Rightarrow \\ $$$${u}^{\mathrm{3}} \:+{v}^{\mathrm{3}} \:+\mathrm{3}{uv}\left({u}+{v}\right)−\frac{{y}^{\mathrm{2}} }{\mathrm{3}}\left({u}+{v}\right)\:−\frac{\mathrm{29}{y}^{\mathrm{3}} }{\mathrm{27}}\:=\mathrm{0}\:\Rightarrow \\ $$$${u}^{\mathrm{3}} \:+{v}^{\mathrm{3}} −\frac{\mathrm{29}{y}^{\mathrm{3}} }{\mathrm{27}}\:+\left({u}+{v}\right)\left(\mathrm{3}{uv}−\frac{{y}^{\mathrm{2}} }{\mathrm{3}}\right)\:=\mathrm{0}\:\Rightarrow{u}^{\mathrm{3}} \:+{v}^{\mathrm{3}} =\frac{\mathrm{29}{y}^{\mathrm{3}} }{\mathrm{27}}\:{and} \\ $$$${uv}\:=\frac{{y}^{\mathrm{2}} }{\mathrm{9}}\:\Rightarrow{u}^{\mathrm{3}} .{v}^{\mathrm{3}} \:=\frac{{y}^{\mathrm{6}} }{\mathrm{9}^{\mathrm{3}} }\:\:{so}\:{u}^{\mathrm{3}} \:{and}\:{v}^{\mathrm{3}} \:{are}\:{roots}\:{of}\:{the}\:{equation} \\ $$$${X}^{\mathrm{2}} −\frac{\mathrm{29}{y}^{\mathrm{3}} }{\mathrm{27}}\:{X}+\frac{{y}^{\mathrm{6}} }{\mathrm{9}^{\mathrm{3}} }\:=\mathrm{0} \\ $$$$\Delta\:=\left(−\frac{\mathrm{29}}{\mathrm{27}}{y}^{\mathrm{3}} \right)^{\mathrm{2}} −\frac{\mathrm{4}{y}^{\mathrm{6}} }{\mathrm{9}^{\mathrm{3}} }\:=\left(\left(\frac{\mathrm{29}}{\mathrm{27}}\right)^{\mathrm{2}} −\frac{\mathrm{4}}{\mathrm{9}^{\mathrm{3}} }\right){y}^{\mathrm{6}} \\ $$$${X}_{\mathrm{1}} =\frac{\frac{\mathrm{29}}{\mathrm{27}}{y}^{\mathrm{3}} \:+\sqrt{\Delta}}{\mathrm{2}}\:\:\:{and}\:{X}_{\mathrm{2}} =\frac{\frac{\mathrm{29}}{\mathrm{27}}{y}^{\mathrm{3}} −\sqrt{\Delta}}{\mathrm{2}}\:\Rightarrow{u}\:=^{\mathrm{3}} \sqrt{{X}_{\mathrm{1}} }\:\:{and}\:{v}=^{\mathrm{3}} \sqrt{{X}_{\mathrm{2}} } \\ $$$${u}=^{\mathrm{3}} \sqrt{{X}_{\mathrm{2}} }\:{and}\:{v}=^{\mathrm{3}} \sqrt{{X}_{\mathrm{1}} } \\ $$$${if}\:{we}\:{take}\:{t}\:=^{\mathrm{3}} \sqrt{{X}_{\mathrm{1}} }\:+^{\mathrm{3}} \sqrt{{X}_{\mathrm{2}} }\:\Rightarrow{x}\:={f}^{−\mathrm{1}} \left({y}\right)\:={t}+\frac{{y}}{\mathrm{3}} \\ $$$$=^{\mathrm{3}} \sqrt{\frac{\left.\frac{\mathrm{29}}{\mathrm{27}}{y}^{\mathrm{3}} +\sqrt{\left(\left(\frac{\mathrm{29}}{\mathrm{27}}\right)^{\mathrm{2}} −\frac{\mathrm{4}}{\mathrm{9}^{\mathrm{3}} }\right.}\right){y}^{\mathrm{3}} }{\mathrm{2}}}\:+^{\mathrm{3}} \sqrt{\frac{\frac{\mathrm{29}}{\mathrm{27}}{y}^{\mathrm{3}} −\sqrt{\left(\left(\frac{\mathrm{29}}{\mathrm{27}}\right)^{\mathrm{2}} −\frac{\mathrm{4}}{\mathrm{9}^{\mathrm{3}} }\right)}{y}^{\mathrm{3}} }{\mathrm{2}}} \\ $$$$+\frac{{y}}{\mathrm{3}} \\ $$
Commented by gunawan last updated on 13/Sep/19
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by mathmax by abdo last updated on 13/Sep/19
you are welcome sir.
$${you}\:{are}\:{welcome}\:{sir}. \\ $$
Answered by MJS last updated on 13/Sep/19
x^3 −yx^2 −y=0 x=t+(y/3) t^3 −(y^2 /3)t−(y/(27))(2y^2 +27)=0 D=(p^3 /(27))+(q^2 /4)=(y^2 /(108))(4y^2 +27)≥0∀y∈R ⇒ Cardano′s method t=(((y/(54))(2y^2 +27−3(√(12y^2 +81))))^(1/3) +(((y/(54))(2y^2 +27+3(√(12y^2 +81))))^(1/3) x=t+(y/3) ⇒ y^(−1) =(x/3)+(((x/(54))(2x^2 +27−3(√(12x^2 +81))))^(1/3) +(((x/(54))(2x^2 +27+3(√(12x^2 +81))))^(1/3)
$${x}^{\mathrm{3}} −{yx}^{\mathrm{2}} −{y}=\mathrm{0} \\ $$$${x}={t}+\frac{{y}}{\mathrm{3}} \\ $$$${t}^{\mathrm{3}} −\frac{{y}^{\mathrm{2}} }{\mathrm{3}}{t}−\frac{{y}}{\mathrm{27}}\left(\mathrm{2}{y}^{\mathrm{2}} +\mathrm{27}\right)=\mathrm{0} \\ $$$${D}=\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}=\frac{{y}^{\mathrm{2}} }{\mathrm{108}}\left(\mathrm{4}{y}^{\mathrm{2}} +\mathrm{27}\right)\geqslant\mathrm{0}\forall{y}\in\mathbb{R} \\ $$$$\Rightarrow\:\mathrm{Cardano}'\mathrm{s}\:\mathrm{method} \\ $$$${t}=\sqrt[{\mathrm{3}}]{\frac{{y}}{\mathrm{54}}\left(\mathrm{2}{y}^{\mathrm{2}} +\mathrm{27}−\mathrm{3}\sqrt{\mathrm{12}{y}^{\mathrm{2}} +\mathrm{81}}\right.}+\sqrt[{\mathrm{3}}]{\frac{{y}}{\mathrm{54}}\left(\mathrm{2}{y}^{\mathrm{2}} +\mathrm{27}+\mathrm{3}\sqrt{\mathrm{12}{y}^{\mathrm{2}} +\mathrm{81}}\right.} \\ $$$${x}={t}+\frac{{y}}{\mathrm{3}} \\ $$$$\Rightarrow \\ $$$${y}^{−\mathrm{1}} =\frac{{x}}{\mathrm{3}}+\sqrt[{\mathrm{3}}]{\frac{{x}}{\mathrm{54}}\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{27}−\mathrm{3}\sqrt{\mathrm{12}{x}^{\mathrm{2}} +\mathrm{81}}\right.}+\sqrt[{\mathrm{3}}]{\frac{{x}}{\mathrm{54}}\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{27}+\mathrm{3}\sqrt{\mathrm{12}{x}^{\mathrm{2}} +\mathrm{81}}\right.} \\ $$
Commented by MJS last updated on 13/Sep/19
x^3 +px+q=0 if D=(p^3 /(27))+(q^2 /4)>0 ⇒ Cardano′s method if D=(p^3 /(27))+(q^2 /4)<0 ⇒ trigonometric method look at qu. 68141, the formulas are there
$${x}^{\mathrm{3}} +{px}+{q}=\mathrm{0} \\ $$$$\mathrm{if}\:{D}=\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}>\mathrm{0}\:\Rightarrow\:\mathrm{Cardano}'\mathrm{s}\:\mathrm{method} \\ $$$$\mathrm{if}\:{D}=\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}<\mathrm{0}\:\Rightarrow\:\mathrm{trigonometric}\:\mathrm{method} \\ $$$$\mathrm{look}\:\mathrm{at}\:\mathrm{qu}.\:\mathrm{68141},\:\mathrm{the}\:\mathrm{formulas}\:\mathrm{are}\:\mathrm{there} \\ $$
Commented by gunawan last updated on 13/Sep/19
Nice Sir What is Cardano′s method ?
$$\mathrm{Nice}\:\mathrm{Sir} \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{Cardano}'\mathrm{s}\:\mathrm{method}\:? \\ $$

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