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Question Number 149817 by ajfour last updated on 07/Aug/21
x^3 −x=c let x=t+h t^3 +3ht^2 +(3h^2 −1)t+h^3 −h−c=0 let t(t^2 +3h^2 −1)=p 3ht^2 +h^3 −h−c=q ⇒ t^2 =((q+c+h−h^3 )/(3h)) p+q=0 (((q+c+h−h^3 )/(3h)))(((q+c+h−h^3 )/(3h))+3h^2 −1)^2 =q^2 ⇒ (q+c+h−h^3 )(q+c−2h+8h^3 )^2 = 27h^3 q^2 let q+c−2h=0 ⇒ 64h^4 (3−h^2 )=27(2h−c)^2 ⇒ 8h^2 (√(3−h^2 ))=3(√3)(2h−c) let h=(√3)sin θ ⇒ 8sin^2 θcos θ=2(√3)sin θ−c 4sin θsin 2θ=2(√3)sin θ−c 2sin θ((√3)−2sin 2θ)=c ......
$$\:\:\:{x}^{\mathrm{3}} −{x}={c} \\ $$$$\:\:{let}\:\:{x}={t}+{h} \\ $$$${t}^{\mathrm{3}} +\mathrm{3}{ht}^{\mathrm{2}} +\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right){t}+{h}^{\mathrm{3}} −{h}−{c}=\mathrm{0} \\ $$$${let}\:\:{t}\left({t}^{\mathrm{2}} +\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right)={p} \\ $$$$\mathrm{3}{ht}^{\mathrm{2}} +{h}^{\mathrm{3}} −{h}−{c}={q} \\ $$$$\Rightarrow\:\:{t}^{\mathrm{2}} =\frac{{q}+{c}+{h}−{h}^{\mathrm{3}} }{\mathrm{3}{h}} \\ $$$${p}+{q}=\mathrm{0} \\ $$$$\left(\frac{{q}+{c}+{h}−{h}^{\mathrm{3}} }{\mathrm{3}{h}}\right)\left(\frac{{q}+{c}+{h}−{h}^{\mathrm{3}} }{\mathrm{3}{h}}+\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:={q}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\left({q}+{c}+{h}−{h}^{\mathrm{3}} \right)\left({q}+{c}−\mathrm{2}{h}+\mathrm{8}{h}^{\mathrm{3}} \right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:=\:\mathrm{27}{h}^{\mathrm{3}} {q}^{\mathrm{2}} \\ $$$${let}\:\:{q}+{c}−\mathrm{2}{h}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\mathrm{64}{h}^{\mathrm{4}} \left(\mathrm{3}−{h}^{\mathrm{2}} \right)=\mathrm{27}\left(\mathrm{2}{h}−{c}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{8}{h}^{\mathrm{2}} \sqrt{\mathrm{3}−{h}^{\mathrm{2}} }=\mathrm{3}\sqrt{\mathrm{3}}\left(\mathrm{2}{h}−{c}\right) \\ $$$${let}\:\:{h}=\sqrt{\mathrm{3}}\mathrm{sin}\:\theta \\ $$$$\Rightarrow\:\:\mathrm{8sin}\:^{\mathrm{2}} \theta\mathrm{cos}\:\theta=\mathrm{2}\sqrt{\mathrm{3}}\mathrm{sin}\:\theta−{c} \\ $$$$\mathrm{4sin}\:\theta\mathrm{sin}\:\mathrm{2}\theta=\mathrm{2}\sqrt{\mathrm{3}}\mathrm{sin}\:\theta−{c} \\ $$$$\mathrm{2sin}\:\theta\left(\sqrt{\mathrm{3}}−\mathrm{2sin}\:\mathrm{2}\theta\right)={c} \\ $$$$…… \\ $$

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