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Question Number 218165 by hardmath last updated on 31/Mar/25
λ > 0 x , y , z ∈ C Solve the system: { ((xy = z^2 + 2λz − λx − λy)),((yz = x^2 + 2λx − λy − λz )),((zx = y^2 + 2λy − λz − λx)) :}
$$\lambda\:>\:\mathrm{0} \\ $$$$\mathrm{x}\:,\:\mathrm{y}\:,\:\mathrm{z}\:\in\:\mathrm{C} \\ $$$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{system}:\:\:\:\begin{cases}{\mathrm{xy}\:=\:\mathrm{z}^{\mathrm{2}} \:+\:\mathrm{2}\lambda\mathrm{z}\:−\:\lambda\mathrm{x}\:−\:\lambda\mathrm{y}}\\{\mathrm{yz}\:=\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{2}\lambda\mathrm{x}\:−\:\lambda\mathrm{y}\:−\:\lambda\mathrm{z}\:\:\:\:\:\:\:\:}\\{\mathrm{zx}\:=\:\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{2}\lambda\mathrm{y}\:−\:\lambda\mathrm{z}\:−\:\lambda\mathrm{x}}\end{cases} \\ $$
Answered by mr W last updated on 01/Apr/25
if x=y=z, then x=y=z=any otherwise, i.e. if x≠y≠z, (i)+(ii)+(iii): xy+yz+zx=x^2 +y^2 +z^2 (ii)−(i): −y(x−z)=x^2 −z^2 +3λ(x−z) (x−z)(x+y+z+3λ)=0 since x≠z, ⇒x+y+z=−3λ (x+y+z)^2 =x^2 +y^2 +z^2 +2(xy+yz+zx) (−3λ)^2 =3(xy+yz+zx) ⇒xy+yz+zx=3λ^2 xyz=z^3 +2λz^2 −λzx−λyz ...(I) xyz=x^3 +2λx^2 −λxy−λzx ...(II) xyz=y^3 +2λy^2 −λyz−λxy ...(III) 2xyz=x^3 +y^3 +z^3 +2λ(x^2 +y^2 +z^2 )−2λ(xy+yz+zx) 2xyz=x^3 +y^3 +z^3 (x+y+z)^3 =x^3 +y^3 +z^3 −3xyz+3(x+y+z)(xy+yz+zx) −27λ^3 =2xyz−3xyz+3(−3λ)(3λ^2 ) −27λ^3 =−xyz−27λ^3 ⇒xyz=0 ⇒x,y,z are roots of r^3 +3λr^2 +3λ^2 r=0 r(r^2 +3λr+3λ^2 )=0 ⇒r=0, ((−(√3)((√3)±i)λ)/2) ⇒(x,y,z)=(0,− (((√3)((√3)+i)λ)/2),−(((√3)((√3)−i)λ)/2))
$${if}\:{x}={y}={z},\:{then}\:{x}={y}={z}={any} \\ $$$${otherwise},\:{i}.{e}.\:{if}\:{x}\neq{y}\neq{z}, \\ $$$$\left({i}\right)+\left({ii}\right)+\left({iii}\right): \\ $$$${xy}+{yz}+{zx}={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \\ $$$$\left({ii}\right)−\left({i}\right): \\ $$$$−{y}\left({x}−{z}\right)={x}^{\mathrm{2}} −{z}^{\mathrm{2}} +\mathrm{3}\lambda\left({x}−{z}\right) \\ $$$$\left({x}−{z}\right)\left({x}+{y}+{z}+\mathrm{3}\lambda\right)=\mathrm{0} \\ $$$${since}\:{x}\neq{z}, \\ $$$$\Rightarrow{x}+{y}+{z}=−\mathrm{3}\lambda \\ $$$$\left({x}+{y}+{z}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +\mathrm{2}\left({xy}+{yz}+{zx}\right) \\ $$$$\left(−\mathrm{3}\lambda\right)^{\mathrm{2}} =\mathrm{3}\left({xy}+{yz}+{zx}\right) \\ $$$$\Rightarrow{xy}+{yz}+{zx}=\mathrm{3}\lambda^{\mathrm{2}} \\ $$$${xyz}={z}^{\mathrm{3}} +\mathrm{2}\lambda{z}^{\mathrm{2}} −\lambda{zx}−\lambda{yz}\:\:\:…\left({I}\right) \\ $$$${xyz}={x}^{\mathrm{3}} +\mathrm{2}\lambda{x}^{\mathrm{2}} −\lambda{xy}−\lambda{zx}\:\:\:…\left({II}\right) \\ $$$${xyz}={y}^{\mathrm{3}} +\mathrm{2}\lambda{y}^{\mathrm{2}} −\lambda{yz}−\lambda{xy}\:\:\:…\left({III}\right) \\ $$$$\mathrm{2}{xyz}={x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} +\mathrm{2}\lambda\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)−\mathrm{2}\lambda\left({xy}+{yz}+{zx}\right) \\ $$$$\mathrm{2}{xyz}={x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} \\ $$$$\left({x}+{y}+{z}\right)^{\mathrm{3}} ={x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} −\mathrm{3}{xyz}+\mathrm{3}\left({x}+{y}+{z}\right)\left({xy}+{yz}+{zx}\right) \\ $$$$−\mathrm{27}\lambda^{\mathrm{3}} =\mathrm{2}{xyz}−\mathrm{3}{xyz}+\mathrm{3}\left(−\mathrm{3}\lambda\right)\left(\mathrm{3}\lambda^{\mathrm{2}} \right) \\ $$$$−\mathrm{27}\lambda^{\mathrm{3}} =−{xyz}−\mathrm{27}\lambda^{\mathrm{3}} \\ $$$$\Rightarrow{xyz}=\mathrm{0} \\ $$$$\Rightarrow{x},{y},{z}\:{are}\:{roots}\:{of} \\ $$$${r}^{\mathrm{3}} +\mathrm{3}\lambda{r}^{\mathrm{2}} +\mathrm{3}\lambda^{\mathrm{2}} {r}=\mathrm{0} \\ $$$${r}\left({r}^{\mathrm{2}} +\mathrm{3}\lambda{r}+\mathrm{3}\lambda^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow{r}=\mathrm{0},\:\frac{−\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{3}}\pm{i}\right)\lambda}{\mathrm{2}} \\ $$$$\Rightarrow\left({x},{y},{z}\right)=\left(\mathrm{0},−\:\frac{\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{3}}+{i}\right)\lambda}{\mathrm{2}},−\frac{\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{3}}−{i}\right)\lambda}{\mathrm{2}}\right) \\ $$
Commented by hardmath last updated on 31/Mar/25
thank you very much dear professor
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{dear}\:\mathrm{professor} \\ $$
Answered by Rasheed.Sindhi last updated on 01/Apr/25
Solve the system: { ((xy = z^2 + 2λz − λx − λy)),((yz = x^2 + 2λx − λy − λz )),((zx = y^2 + 2λy − λz − λx)) :} Adding: xy+yz+zx=x^2 +y^2 +z^2 x^2 +y^2 +z^2 −xy−yz−zx=0 2(x^2 +y^2 +z^2 −xy−yz−zx)=0 (x−y)^2 +(y−z)^2 +(z−x)^2 =0 {: ((x−y=0⇒x=y)),((y−z=0⇒y=z)) }⇒x=y=z x=y=z=a ; a∈R
$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{system}:\:\:\:\begin{cases}{\mathrm{xy}\:=\:\mathrm{z}^{\mathrm{2}} \:+\:\mathrm{2}\lambda\mathrm{z}\:−\:\lambda\mathrm{x}\:−\:\lambda\mathrm{y}}\\{\mathrm{yz}\:=\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{2}\lambda\mathrm{x}\:−\:\lambda\mathrm{y}\:−\:\lambda\mathrm{z}\:\:\:\:\:\:\:\:}\\{\mathrm{zx}\:=\:\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{2}\lambda\mathrm{y}\:−\:\lambda\mathrm{z}\:−\:\lambda\mathrm{x}}\end{cases} \\ $$$${Adding}: \\ $$$${xy}+{yz}+{zx}={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} −{xy}−{yz}−{zx}=\mathrm{0} \\ $$$$\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} −{xy}−{yz}−{zx}\right)=\mathrm{0} \\ $$$$\left({x}−{y}\right)^{\mathrm{2}} +\left({y}−{z}\right)^{\mathrm{2}} +\left({z}−{x}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\left.\begin{matrix}{{x}−{y}=\mathrm{0}\Rightarrow{x}={y}}\\{{y}−{z}=\mathrm{0}\Rightarrow{y}={z}}\end{matrix}\right\}\Rightarrow{x}={y}={z} \\ $$$${x}={y}={z}={a}\:;\:{a}\in\mathbb{R} \\ $$$$ \\ $$
Commented by mr W last updated on 01/Apr/25
since x,y,z ∈C, from (x−y)^2 +(y−z)^2 +(z−x)^2 =0 it doesn′t only follow x−y=0, y−z=0, z−x=0
$${since}\:{x},{y},{z}\:\in{C},\:{from} \\ $$$$\left({x}−{y}\right)^{\mathrm{2}} +\left({y}−{z}\right)^{\mathrm{2}} +\left({z}−{x}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${it}\:{doesn}'{t}\:{only}\:{follow} \\ $$$${x}−{y}=\mathrm{0},\:{y}−{z}=\mathrm{0},\:{z}−{x}=\mathrm{0}\: \\ $$
Commented by Rasheed.Sindhi last updated on 01/Apr/25
Right sir! This was only for real numbers.
$${Right}\:{sir}!\:{This}\:{was}\:{only}\:{for}\:{real}\:{numbers}. \\ $$
Commented by mr W last updated on 01/Apr/25
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