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Question Number 155928 by cortano last updated on 06/Oct/21

Answered by mr W last updated on 06/Oct/21

$${(u+v)}^3=u^3+v^3+3uv(u+v)$$$${(u+v)}^3-3uv(u+v)-(u^3+v^3)=0$$$$x^3-3x+4=0$$$$\Rightarrow x=u+v$$$$\Rightarrow uv=1\Rightarrow u^3{v}^3=1$$$$\Rightarrow u^3+v^3=-4$$$$u^3,v^{3}arerootsof{z}^2+4z+1=0$$$$\Rightarrow u^3,v^3=z=-2\pm \sqrt{3}$$$$\Rightarrow u=\sqrt[3]{-2+\sqrt{3}},v=\sqrt[3]{-2-\sqrt{3}}$$$$\Rightarrow x=u+v=\sqrt[3]{-2+\sqrt{3}}+\sqrt[3]{-2-\sqrt{3}}$$$$\Rightarrow x=-\sqrt[3]{2-\sqrt{3}}-\sqrt[3]{2+\sqrt{3}}$$

Commented by cortano last updated on 06/Oct/21

$$\mathrm{cardano}.\mathrm{yes}.$$

Commented by mr W last updated on 06/Oct/21

$$yes,cortanoappliescardano!$$

Commented by cortano last updated on 06/Oct/21

$$\mathrm{hahahaha}$$

Commented by Tawa11 last updated on 06/Oct/21

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

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