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Question Number 128660 by help last updated on 09/Jan/21

Commented by harckinwunmy last updated on 09/Jan/21

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Answered by liberty last updated on 09/Jan/21

$$\mathrm{let}\frac{\mathrm{x}}{\mathrm{y}}=\mathrm{u}\wedge \frac{\mathrm{y}}{\mathrm{x}}=\mathrm{v}\Rightarrow \mathrm{u}.\mathrm{v}=1\mathrm{and}\mathrm{u}+\mathrm{v}=1$$$$\iff {(2\mathrm{u}+{\mathrm{v}}^3)}^3+{(2\mathrm{v}+{\mathrm{u}}^3)}^3=(\ast )$$$$\mathrm{consider}(\ast ):{\mathrm{a}}^3+{\mathrm{b}}^3={(\mathrm{a}+\mathrm{b})}^3-3\mathrm{ab}(\mathrm{a}+\mathrm{b})$$$$\mathrm{then}{(2\mathrm{u}+{\mathrm{v}}^3+2\mathrm{v}+{\mathrm{u}}^3)}^3-3(2\mathrm{u}+{\mathrm{v}}^3+2\mathrm{v}+{\mathrm{u}}^3)(2\mathrm{u}+{\mathrm{v}}^3)(2\mathrm{v}+{\mathrm{u}}^3)$$$${(2(\mathrm{u}+\mathrm{v})+{(\mathrm{u}+\mathrm{v})}^3-3\mathrm{uv}(\mathrm{u}+\mathrm{v}))}^3-3(2(\mathrm{u}+\mathrm{v})+{(\mathrm{u}+\mathrm{v})}^3-3\mathrm{uv}(\mathrm{u}+\mathrm{v})(2\mathrm{u}+{\mathrm{v}}^3)(2\mathrm{v}+{\mathrm{u}}^3)$$$$={(2.1+1-3.1.1)}^3-3(2.1+1-3.1.1)(4\mathrm{uv}+{2\mathrm{u}}^4+{2\mathrm{v}}^4+{\left(\mathrm{uv}\right)}^3)$$$$={\left(0\right)}^3-3\left(0\right)(4+1+2({\mathrm{u}}^4+{\mathrm{v}}^4))=0$$

Commented by bemath last updated on 09/Jan/21

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