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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

let (x/y) = u ∧ (y/x) = v ⇒u.v=1and u+v=1 ⇔(2u+v^3 )^3 +(2v+u^3 )^3 = (∗) consider (∗):a^3 +b^3 = (a+b)^3 −3ab(a+b) then (2u+v^3 +2v+u^3 )^3 −3(2u+v^3 +2v+u^3 )(2u+v^3 )(2v+u^3 ) (2(u+v)+(u+v)^3 −3uv(u+v))^3 −3(2(u+v)+(u+v)^3 −3uv(u+v)(2u+v^3 )(2v+u^3 ) =(2.1+1−3.1.1)^3 −3(2.1+1−3.1.1)(4uv+2u^4 +2v^4 +(uv)^3 ) =(0)^3 −3(0)(4+1+2(u^4 +v^4 )) = 0

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

Commented by bemath last updated on 09/Jan/21

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