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Question Number 201679 by cherokeesay last updated on 10/Dec/23

Answered by Rasheed.Sindhi last updated on 10/Dec/23

$$\sqrt[6]{1-\sqrt{x^2+1}}+\sqrt[6]{\sqrt{x^2+1}-1}=1$$$$a+b=1\Rightarrow b=1-a$$$$a^6+b^6=1-\sqrt{x^2+1}+\sqrt{x^2+1}-1=0$$$$(a^2+b^2)(a^4-a^2{b}^2+b^4)=0$$$$a^2+b^2=0\mid a^4-a^2{b}^2+b^4=0^{★}$$$${(a+b)}^2-2ab=0$$$$1-2ab=0$$$$1-2a(1-a)=0$$$$2a^2-2a+1=0$$$$a=\frac{2\pm \sqrt{4-8}}{4}=\frac{1\pm i}{2}$$$$\sqrt[6]{1-\sqrt{x^2+1}}=\frac{1\pm i}{2}$$$$1-\sqrt{x^2+1}={\left(\frac{1\pm i}{2}\right)}^6$$$$\sqrt{x^2+1}=1-{\left(\frac{1\pm i}{2}\right)}^6$$$$x^2+1={(1-{\left(\frac{1\pm i}{2}\right)}^6)}^2$$$$x=\pm \sqrt{{(1-{\left(\frac{1\pm i}{2}\right)}^6)}^2-1}$$$$=\pm \sqrt{1-2{\left(\frac{1\pm i}{2}\right)}^6+{\left(\frac{1\pm i}{2}\right)}^{12}-1}$$$$=\pm \sqrt{-2{\left(\frac{1\pm i}{2}\right)}^6+{\left(\frac{1\pm i}{2}\right)}^{12}}$$$$=\pm {\left(\frac{1\pm i}{2}\right)}^6\sqrt{-2+{\left(\frac{1\pm i}{2}\right)}^6}$$$$....$$$$...$$

Answered by Rasheed.Sindhi last updated on 10/Dec/23

$$1-\sqrt{x^2+1}=a\left(say\right)$$$$\sqrt[6]{a}+\sqrt[6]{-a}=1$$$${(\sqrt[6]{a}+\sqrt[6]{-a})}^2=1^2$$$$a^{1/3}+2a^{1/6}{(-a)}^{1/6}+{(-a)}^{1/3}=1$$$$a^{1/3}+{(-a)}^{1/3}=1-2a^{1/6}{(-a)}^{1/6}$$$${(a^{1/3}+{(-a)}^{1/3})}^3={(1-2a^{1/6}{(-a)}^{1/6})}^3$$$$\{a+(-a)\}+3\left(a^{1/3}\right){(-a)}^{1/3}\{a^{1/3}+{(-a)}^{1/3}\}$$$$=.....$$$$$$

Answered by mr W last updated on 10/Dec/23

$$letu=\sqrt[6]{1-\sqrt{x^2+1}}$$$$\sqrt[6]{\sqrt{x^2+1}-1}=\sqrt[6]{(-1)(1-\sqrt{x^2+1})}=\sqrt[3]{i}u$$$$u+\sqrt[3]{i}u=1$$$$u(1+\sqrt[3]{i})=1$$$$u=\frac{1}{1+\sqrt[3]{i}}=\sqrt[6]{1-\sqrt{x^2+1}}$$$$\text{Missing left or extra right}$$$$\text{Missing left or extra right}$$

Answered by Frix last updated on 10/Dec/23

$$z^{\frac{1}{6}}+{(-z)}^{\frac{1}{6}}=1$$$$z=r{\mathrm{e}}^{\mathrm{i}\theta }\wedge -z=r{\mathrm{e}}^{\mathrm{i}(\theta -\pi )}$$$$z^{\frac{1}{6}}+{(-z)}^{\frac{1}{6}}\in \mathbb{R}\Rightarrow \theta =-(\theta -\pi )\Rightarrow \theta =\frac{\pi }{2}$$$$\Rightarrow z=r{\mathrm{e}}^{\mathrm{i}\frac{\pi }{2}}\wedge -z=r{\mathrm{e}}^{-\mathrm{i}\frac{\pi }{2}}$$$${\left(r{\mathrm{e}}^{\mathrm{i}\frac{\pi }{2}}\right)}^{\frac{1}{6}}+{\left(r{\mathrm{e}}^{-\mathrm{i}\frac{\pi }{2}}\right)}^{\frac{1}{6}}=1$$$$r^{\frac{1}{6}}({\mathrm{e}}^{\mathrm{i}\frac{\pi }{12}}+{\mathrm{e}}^{-\mathrm{i}\frac{\pi }{12}})=1$$$$r^{\frac{1}{6}}\frac{\sqrt{6}+\sqrt{2}}{2}=1$$$$r^{\frac{1}{6}}=\frac{\sqrt{6}-\sqrt{2}}{2}$$$$r=26-15\sqrt{3}$$$$z=(26-15\sqrt{3})\mathrm{i}$$$$(26-15\sqrt{3})\mathrm{i}=\pm (1-\sqrt{x^2+1})$$$$\mathrm{This}\mathrm{can}\mathrm{be}\mathrm{solved}\mathrm{exactly}...$$

Answered by witcher3 last updated on 10/Dec/23

$$\mathrm{no}\mathrm{real}\mathrm{solution}$$$${(1-\sqrt{{\mathrm{x}}^2+1})}^{\frac{1}{6}}={\mathrm{e}}^{\frac{\mathrm{i}\pi }{6}}{(\sqrt{{\mathrm{x}}^2+1}-1)}^{\frac{1}{6}}+{(\sqrt{{\mathrm{x}}^2+1}-1)}^{\frac{1}{6}}=1$$$$\sqrt{{\mathrm{x}}^2+1}-1={\left(\frac{1}{{\mathrm{e}}^{\frac{\mathrm{i}\pi }{6}}+1}\right)}^6=\frac{{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{2}}}{{\left(2\cos \left(\frac{\pi }{12}\right)\right)}^6}=-\frac{\mathrm{i}}{{32\cos }^6\left(\frac{\pi }{12}\right)}=\mathrm{a}$$$${\mathrm{x}}^2+1={(\mathrm{a}+1)}^2$$$${\mathrm{x}}^2={\mathrm{a}}^2+2\mathrm{a}$$$$\mathrm{x}=\underset{-}{+}\sqrt{{\mathrm{a}}^2+2\mathrm{a}}$$$$\mathrm{a}=-\frac{\mathrm{i}}{{32\cos }^6\left(\frac{\pi }{12}\right)}$$

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