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

((x + 2))^(1/3) − ((x − 3))^(1/3) > (1/2)

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

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

t^3 =x−3⇒t^3 +5=x+2⇒ ((t^3 +5))^(1/3) ⟩t+(1/2)⇒power 3 t^3 +5⟩t^3 +(3/2)t^2 +(3/4)t+(1/8)⇒ 12t^2 +6t−39⟨0⇒4t^2 +2t−13⟨0⇒ Δ=4+208=212 t_(1,2) =((−2±(√(212)))/8)=((−1±(√(53)))/4) x_(1,2) =t^3 +3=(((−1±(√(53)))/4))^3 +3 now we can solve easily

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

Commented byTawa1 last updated on 18/Jan/19

God bless you sir

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

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