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Question Number 155154 by mathdanisur last updated on 26/Sep/21

$$\mathrm{let}\mathrm{n}\in {\mathbb{N}}^{+}\mathrm{solve}\mathrm{for}\mathrm{real}\mathrm{numbers}:$$$${\mathrm{x}}^{3\mathbf{n}}-{\mathrm{y}}^{2\mathbf{n}}={\mathrm{y}}^{3\mathbf{n}}-{\mathrm{x}}^{2\mathbf{n}}=4$$

Answered by MJS_new last updated on 26/Sep/21

$$x^{3n}-y^{2n}=y^{3n}-x^{2n}\Rightarrow y=x$$$$x^{3n}-x^{2n}-4=0$$$$(x^n-2)(x^{2n}+x^n+2)=0$$$$\Rightarrow x=y=2^{1/n}$$

Commented by mathdanisur last updated on 26/Sep/21

$$\mathrm{Very}\mathrm{nice},\mathrm{thank}\mathrm{you}\mathrm{my}\mathrm{dear}$$

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