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Question Number 202500 by Calculusboy last updated on 28/Dec/23

Answered by Frix last updated on 28/Dec/23

$$\mathrm{Obviously}x=1$$$$\sqrt[2015]{1+3-3}+\sqrt[2015]{-1-3+5}=1+1=2$$

Answered by Rasheed.Sindhi last updated on 28/Dec/23

$$\sqrt[2015]{x^3+3x-3}+\sqrt[2015]{-x^3-3x+3+2}=2$$$$\sqrt[2015]{x^3+3x-3}+\sqrt[2015]{-(x^3+3x-3)+2}=2$$$$▸\sqrt[2015]{x^3+3x-3}=a\Rightarrow a^{2015}=x^3+3x-3$$$$▸\sqrt[2015]{-(x^3+3x-3)+2}=\sqrt[2015]{2-a}$$$$a+\sqrt[2015]{2-a}=2$$$$\sqrt[2015]{2-a}=2-a$$$$2-a={(2-a)}^{2015}$$$${(2-a)}^{2015-1}=1$$$${(2-a)}^{2014}=1^{2014}$$$$2-a=1$$$$a=1$$$$\sqrt[2015]{x^3+3x-3}=1$$$$x^3+3x-3=1$$$$x^3+3x-4=0$$$$(x-1)(x^2+x+4)=0$$$$x=1\mid x=\frac{-1\pm \sqrt{1-16}}{2}$$$$x=1\mid x=\frac{-1\pm i\sqrt{15}}{2}$$

Commented by Rasheed.Sindhi last updated on 28/Dec/23

$$Erraneousanswer!$$

Commented by Frix last updated on 28/Dec/23

$$\mathrm{I}\mathrm{think}\mathrm{we}\mathrm{need}\sqrt{1}+\sqrt{1}=2\Rightarrow$$$$x^3+3x-4=1$$$$\mathrm{and}\mathrm{your}\mathrm{solution}\mathrm{is}\mathrm{right}.$$$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{think}\mathrm{solutions}\mathrm{for}$$$$\sqrt[n]{r{\mathrm{e}}^{\mathrm{i}\theta }}+\sqrt[n]{2-r{\mathrm{e}}^{\mathrm{i}\theta }}\in \mathbb{R}$$$$\mathrm{exist}\mathrm{when}r{\mathrm{e}}^{\mathrm{i}\theta }\notin \mathbb{R}$$

Commented by esmaeil last updated on 28/Dec/23

$$if(\sqrt[2015]{x^3+3x-3}=a)\to$$$$\sqrt[2015]{2-(x^3+3x-3)}=\sqrt[2015]{2-a^{2015}}$$$$\ne \sqrt[2015]{2-a}$$

Commented by Frix last updated on 28/Dec/23

$$\mathrm{But}\mathrm{if}{x}^3+3x-3=a\Rightarrow \sqrt[2015]{a}+\sqrt[2015]{2-a}$$

Commented by esmaeil last updated on 28/Dec/23

$$yes$$$$butit{dosn}^{\prime }thelpustosolvethat.$$

Commented by Frix last updated on 28/Dec/23

$$\mathrm{It}\mathrm{helps}\mathrm{to}\mathrm{see}{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{no}\mathrm{other}\mathrm{solutions}$$$$\mathrm{besides}a=1.$$

Commented by Calculusboy last updated on 28/Dec/23

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s}\mathit{s}\mathit{i}\mathit{r}$$

Commented by Rasheed.Sindhi last updated on 29/Dec/23

$$\mathbb{T}\mathbf{hanks}\mathbf{sir}\mathrm{for}\mathbf{guidance}!$$

Answered by Rasheed.Sindhi last updated on 28/Dec/23

$$\sqrt[2015]{x^3+3x-3}+\sqrt[2015]{-(x^3+3x-3)+2}=2$$$$\sqrt[2015]{a}+\sqrt[2015]{2-a}=2$$$$A+B=2$$$$A^{2015}+B^{2015}=a+2-a=2$$$$$$

Answered by Rasheed.Sindhi last updated on 29/Dec/23

$$\mathbf{According}\mathbf{to}\mathbf{the}\mathbf{strategy}\mathbf{of}\mathbf{sir}\mathbf{Frix}$$$$\sqrt[2015]{x^3+3x-3}+\sqrt[2015]{-x^3-3x+5}$$$$$$$$\sqrt[2015]{x^3+3x-4+1}+\sqrt[2015]{-(x^3+3x-4)+1}$$$$x^3+3x-4=y$$$$\sqrt[2015]{1+y}+\sqrt[2015]{1-y}=2$$$$Onepossibility:$$$$\sqrt[2015]{1+y}+\sqrt[2015]{1-y}=1+1$$$$\sqrt[2015]{1+y}=1\wedge \sqrt[2015]{1-y}=1$$$$1+y=1^{2015}\wedge 1-y=1^{2015}$$$$1+y=1\wedge 1-y=1\Rightarrow y=0$$$$x^3+3x-4=0$$$$(x-1)(x^2+x+4)=0$$$$x=1,\frac{-1\pm i\sqrt{15}}{2}$$

Commented by mr W last updated on 29/Dec/23

$$ifyouconsiderx\in C,ithinkthere$$$$aremuchmorecomplexrootsthan$$$$\frac{-1\pm i\sqrt{15}}{2},becausey=0istheonly$$$$onerealrootfrom\sqrt[2015]{1+y}+\sqrt[2015]{1-y}=2,$$$$butnottheonlyonecomplexroot$$$$fromit.$$

Commented by Frix last updated on 29/Dec/23

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{think}\sqrt[n]{1+y}+\sqrt[n]{1-y}=2\mathrm{has}\mathrm{complex}$$$$\mathrm{roots}.$$

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