Question Number 211252 by RojaTaniya last updated on 02/Sep/24
Answered by Frix last updated on 02/Sep/24
![[1]×(x−y) ⇒ x^3 −y^3 =39(x−y) [1a] [2]×(y−z) ⇒ y^3 −z^3 =49(y−z) [2a] [3]×(z−x) ⇒ z^3 −x^3 =19(z−x) [3a] [1a]+[2a]+[3a] ⇒ y=3z−2x [1] x^2 −3xz+3z^2 −13=0 [2] 4x^2 −14xz+13z^2 −49=0 [3] x^2 +xz+z^2 −19=0 [3]−[1] ⇒ x=((z^2 +3)/(2z)) [2] z^4 −((64z^2 )/7)+(9/7)=0 z=±(1/( (√7)))∨z=±3 ((x),(y),(z) ) =(√7) (((±11)),((∓19)),((±1)) ) ∨ ((x),(y),(z) ) = (((±2)),((±5)),((±3)) )](https://www.tinkutara.com/question/Q211254.png)
$$\left[\mathrm{1}\right]×\left({x}−{y}\right)\:\Rightarrow\:{x}^{\mathrm{3}} −{y}^{\mathrm{3}} =\mathrm{39}\left({x}−{y}\right)\:\:\left[\mathrm{1}{a}\right] \\ $$$$\left[\mathrm{2}\right]×\left({y}−{z}\right)\:\Rightarrow\:{y}^{\mathrm{3}} −{z}^{\mathrm{3}} =\mathrm{49}\left({y}−{z}\right)\:\:\left[\mathrm{2}{a}\right] \\ $$$$\left[\mathrm{3}\right]×\left({z}−{x}\right)\:\Rightarrow\:{z}^{\mathrm{3}} −{x}^{\mathrm{3}} =\mathrm{19}\left({z}−{x}\right)\:\:\left[\mathrm{3}{a}\right] \\ $$$$\left[\mathrm{1}{a}\right]+\left[\mathrm{2}{a}\right]+\left[\mathrm{3}{a}\right]\:\Rightarrow\:{y}=\mathrm{3}{z}−\mathrm{2}{x} \\ $$$$\left[\mathrm{1}\right]\:\:{x}^{\mathrm{2}} −\mathrm{3}{xz}+\mathrm{3}{z}^{\mathrm{2}} −\mathrm{13}=\mathrm{0} \\ $$$$\left[\mathrm{2}\right]\:\:\mathrm{4}{x}^{\mathrm{2}} −\mathrm{14}{xz}+\mathrm{13}{z}^{\mathrm{2}} −\mathrm{49}=\mathrm{0} \\ $$$$\left[\mathrm{3}\right]\:\:{x}^{\mathrm{2}} +{xz}+{z}^{\mathrm{2}} −\mathrm{19}=\mathrm{0} \\ $$$$\left[\mathrm{3}\right]−\left[\mathrm{1}\right]\:\Rightarrow\:{x}=\frac{{z}^{\mathrm{2}} +\mathrm{3}}{\mathrm{2}{z}} \\ $$$$\left[\mathrm{2}\right]\:\:{z}^{\mathrm{4}} −\frac{\mathrm{64}{z}^{\mathrm{2}} }{\mathrm{7}}+\frac{\mathrm{9}}{\mathrm{7}}=\mathrm{0} \\ $$$${z}=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}}}\vee{z}=\pm\mathrm{3} \\ $$$$\begin{pmatrix}{{x}}\\{{y}}\\{{z}}\end{pmatrix}\:=\sqrt{\mathrm{7}}\begin{pmatrix}{\pm\mathrm{11}}\\{\mp\mathrm{19}}\\{\pm\mathrm{1}}\end{pmatrix}\:\vee\begin{pmatrix}{{x}}\\{{y}}\\{{z}}\end{pmatrix}\:=\begin{pmatrix}{\pm\mathrm{2}}\\{\pm\mathrm{5}}\\{\pm\mathrm{3}}\end{pmatrix} \\ $$