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Question Number 140900 by bramlexs22 last updated on 14/May/21

{ ((x^3 =xyz+1)),((y^3 =xyz+2)),((z^3 =xyz−3)) :}

$$\:\begin{cases}{\mathrm{x}^{\mathrm{3}} =\mathrm{xyz}+\mathrm{1}}\\{\mathrm{y}^{\mathrm{3}} =\mathrm{xyz}+\mathrm{2}}\\{\mathrm{z}^{\mathrm{3}} =\mathrm{xyz}−\mathrm{3}}\end{cases} \\ $$

Commented by bramlexs22 last updated on 14/May/21

Thank all master

$$\mathrm{Thank}\:\mathrm{all}\:\mathrm{master} \\ $$

Answered by EDWIN88 last updated on 14/May/21

(1)×(2)×(3) . let ℓ = xyz ⇒ ℓ^3 = (ℓ+1)(ℓ+2)(ℓ−3) ⇒ ℓ = −(6/7)=xyz ⇒x^3 = (1/7) ⇒x=(1/( (7)^(1/(3 )) )) ⇒y^3 = (8/7)⇒y = (2/( (7)^(1/(3 )) )) ⇒z^3 = −((27)/7) ⇒z =−(3/( (7)^(1/(3 )) ))

$$\left(\mathrm{1}\right)×\left(\mathrm{2}\right)×\left(\mathrm{3}\right)\:.\:\mathrm{let}\:\ell\:=\:\mathrm{xyz} \\ $$$$\Rightarrow\:\ell^{\mathrm{3}} \:=\:\left(\ell+\mathrm{1}\right)\left(\ell+\mathrm{2}\right)\left(\ell−\mathrm{3}\right) \\ $$$$\Rightarrow\:\ell\:=\:−\frac{\mathrm{6}}{\mathrm{7}}=\mathrm{xyz} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{3}} \:=\:\frac{\mathrm{1}}{\mathrm{7}}\:\Rightarrow\mathrm{x}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}\:}]{\mathrm{7}}} \\ $$$$\Rightarrow\mathrm{y}^{\mathrm{3}} =\:\frac{\mathrm{8}}{\mathrm{7}}\Rightarrow\mathrm{y}\:=\:\frac{\mathrm{2}}{\:\sqrt[{\mathrm{3}\:}]{\mathrm{7}}}\: \\ $$$$\Rightarrow\mathrm{z}^{\mathrm{3}} \:=\:−\frac{\mathrm{27}}{\mathrm{7}}\:\Rightarrow\mathrm{z}\:=−\frac{\mathrm{3}}{\:\sqrt[{\mathrm{3}\:}]{\mathrm{7}}}\: \\ $$

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