Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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 )) ))

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com