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Question Number 53108 by Tawa1 last updated on 17/Jan/19

$$\sqrt[3]{\mathrm{x}+2}-\sqrt[3]{\mathrm{x}-3}>\frac{1}{2}$$

Answered by kaivan.ahmadi last updated on 17/Jan/19

$${\mathrm{t}}^3=\mathrm{x}-3\Rightarrow {\mathrm{t}}^3+5=\mathrm{x}+2\Rightarrow$$ $$\sqrt[3]{{\mathrm{t}}^3+5}⟩\mathrm{t}+\frac{1}{2}\Rightarrow \mathrm{power}3$$ $${\mathrm{t}}^3+5⟩{\mathrm{t}}^3+\frac{3}{2}{\mathrm{t}}^2+\frac{3}{4}\mathrm{t}+\frac{1}{8}\Rightarrow$$ $${12\mathrm{t}}^2+6\mathrm{t}-39⟨0\Rightarrow {4\mathrm{t}}^2+2\mathrm{t}-13⟨0\Rightarrow$$ $$\mathrm{\Delta }=4+208=212$$ $${\mathrm{t}}_{1,2}=\frac{-2\pm \sqrt{212}}{8}=\frac{-1\pm \sqrt{53}}{4}$$ $${\mathrm{x}}_{1,2}={\mathrm{t}}^3+3={\left(\frac{-1\pm \sqrt{53}}{4}\right)}^3+3$$ $$\mathrm{now}\mathrm{we}\mathrm{can}\mathrm{solve}\mathrm{easily}$$ $$$$

Commented byTawa1 last updated on 18/Jan/19

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

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