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Question Number 41798 by Tawa1 last updated on 12/Aug/18

$$\mathrm{Solve}:{\mathrm{x}}^3-3\mathrm{x}+4=0$$$$\mathrm{how}\mathrm{can}\mathrm{i}\mathrm{know}\mathrm{when}\mathrm{to}\mathrm{use}:\mathrm{let}\mathrm{x}=\mathrm{y}+\frac{1}{\mathrm{y}}$$

Commented by math khazana by abdo last updated on 13/Aug/18

$$letx=u+v\Rightarrow {(u+v)}^3-3(u+v)+4=0\Rightarrow$$$$u^3+v^3+3uv(u+v)-3(u+v)+4=0\Rightarrow$$$$u^3+v^3+4+(u+v)(3uv-3)=0\Rightarrow$$$$u^3+v^3=-4anduv=1\Rightarrow u^3+v^3=-4and$$$$u^3.v^3=1\Rightarrow u^{3}and{v}^{3}aresolutionoftheequation$$$$X^2+4X+1=0$$$${\mathrm{\Delta }}^{{}^{\prime }}=2^2-1=3\Rightarrow X_1=-2+\sqrt{3}and{X}_2=-2-\sqrt{3}\Rightarrow$$$$\Rightarrow u{=}^3\sqrt{-2+\sqrt{3}}andv{=}^3\sqrt{-2-\sqrt{3}}$$$$\Rightarrow x{=}^3\sqrt{-2+\sqrt{3}}{+}^3\sqrt{-2-\sqrt{3}}$$

Commented by Tawa1 last updated on 13/Aug/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by maxmathsup by imad last updated on 13/Aug/18

$$nevermindsir$$

Commented by soufiane zarik last updated on 13/Aug/18

Commented by Tawa1 last updated on 15/Aug/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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