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Question Number 211157 by RojaTaniya last updated on 30/Aug/24

Commented by Rasheed.Sindhi last updated on 30/Aug/24

(x,y,z)=(1,0,0),(0,1,0),(0,0,1)

$$\left({x},{y},{z}\right)=\left(\mathrm{1},\mathrm{0},\mathrm{0}\right),\left(\mathrm{0},\mathrm{1},\mathrm{0}\right),\left(\mathrm{0},\mathrm{0},\mathrm{1}\right) \\ $$

Answered by Rasheed.Sindhi last updated on 30/Aug/24

{ ((x+y+z=1...(i))),((x^2 y+y^2 z+z^2 x=xy^2 +yz^2 +zx^2 ...(ii))),((x^3 +y^2 +z=y^3 +z^2 +x...(iii))) :} Assuming x,y,z>0 (ii)⇒xy(x−y)+yz(y−z)+zx(z−x)=0 xy(x−y)=0∧yz(y−z)=0∧zx(z−x)=0 x−y=0∧y−z=0∧z−x=0 ⇒x=y=z (i)⇒x=y=z=1/3

$$\begin{cases}{{x}+{y}+{z}=\mathrm{1}...\left({i}\right)}\\{{x}^{\mathrm{2}} {y}+{y}^{\mathrm{2}} {z}+{z}^{\mathrm{2}} {x}={xy}^{\mathrm{2}} +{yz}^{\mathrm{2}} +{zx}^{\mathrm{2}} ...\left({ii}\right)}\\{{x}^{\mathrm{3}} +{y}^{\mathrm{2}} +{z}={y}^{\mathrm{3}} +{z}^{\mathrm{2}} +{x}...\left({iii}\right)}\end{cases} \\ $$$${Assuming}\:{x},{y},{z}>\mathrm{0} \\ $$$$\left({ii}\right)\Rightarrow{xy}\left({x}−{y}\right)+{yz}\left({y}−{z}\right)+{zx}\left({z}−{x}\right)=\mathrm{0} \\ $$$${xy}\left({x}−{y}\right)=\mathrm{0}\wedge{yz}\left({y}−{z}\right)=\mathrm{0}\wedge{zx}\left({z}−{x}\right)=\mathrm{0} \\ $$$${x}−{y}=\mathrm{0}\wedge{y}−{z}=\mathrm{0}\wedge{z}−{x}=\mathrm{0} \\ $$$$\Rightarrow{x}={y}={z} \\ $$$$\left({i}\right)\Rightarrow{x}={y}={z}=\mathrm{1}/\mathrm{3} \\ $$

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