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Question Number 45546 by Tawa1 last updated on 14/Oct/18

$$\mathrm{Simplify}:\sqrt[3]{\sqrt{5}+2}+\sqrt[3]{\sqrt{5}-2}$$

Commented by Meritguide1234 last updated on 14/Oct/18

$$x^3=(\sqrt{5}+2)+(\sqrt{5}-2)+3.x$$$$$$

Answered by Meritguide1234 last updated on 14/Oct/18

$$x^3-3x-2\sqrt{5}=0$$

Commented by Meritguide1234 last updated on 14/Oct/18

$$x(x^2-3)=\sqrt{5}({\left(\sqrt{5}\right)}^2-3)$$$$x=\sqrt{5}\left(uniquesolution\right)$$

Answered by MJS last updated on 14/Oct/18

$$x=\sqrt[3]{2+\sqrt{5}}-\sqrt[3]{2-\sqrt{5}}$$$$x^3=(2+\sqrt{5})-3{\left({(2+\sqrt{5})}^2(2-\sqrt{5})\right)}^{\frac{1}{3}}+3{\left((2+\sqrt{5}){(2-\sqrt{5})}^2\right)}^{\frac{1}{3}}-(2-\sqrt{5})$$$$x^3=2\sqrt{5}-3{\left((2+\sqrt{5})(-1)\right)}^{\frac{1}{3}}+3{\left((2-\sqrt{5})(-1)\right)}^{\frac{1}{3}}$$$$x^3=2\sqrt{5}+3(\sqrt[3]{2+\sqrt{5}}-\sqrt[3]{2-\sqrt{5}})$$$$x^3=2\sqrt{5}+3x$$$$x^3-3x=2\sqrt{5}$$$$\mathrm{in}\mathrm{this}\mathrm{special}\mathrm{case}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easy}\mathrm{to}\mathrm{see}\mathrm{that}{x}_1=\sqrt{5}$$$$\mathrm{because}{\left(\sqrt{5}\right)}^3=5\sqrt{5}$$$$$$$$x^3-3x-2\sqrt{5}=(x-\sqrt{5})(x^2+\sqrt{5}x+2)$$$$x_2,x_3\notin \mathbb{R}$$$$\Rightarrow x=\sqrt{5}$$$$$$$$\mathrm{btw}.:{(\frac{1}{2}+\frac{\sqrt{5}}{2})}^3=2+\sqrt{5}{(-\frac{1}{2}+\frac{\sqrt{5}}{2})}^3=-2+\sqrt{5}$$

Commented by Tawa1 last updated on 14/Oct/18

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

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