Let x, y, z be positive real numbers. Which o | vedclass

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(B) For $x, y, z > 0$,we analyze each condition:$I.$ $x^3+y^3+z^3-3xyz = \frac{1}{2}(x+y+z)((x-y)^2+(y-z)^2+(z-x)^2) = 0$. Since $x+y+z > 0$,this impl

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$I, II, IV$ only

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(B) For $x, y, z > 0$,we analyze each condition:$I.$ $x^3+y^3+z^3-3xyz = \frac{1}{2}(x+y+z)((x-y)^2+(y-z)^2+(z-x)^2) = 0$. Since $x+y+z > 0$,this implies $(x-y)^2+(y-z)^2+(z-x)^2 = 0$,so $x=y=z$.$II.$ $x^3+y^2z+yz^2=3xyz$. Dividing by $xyz$,we get $\frac{x^2}{yz} + \frac{y}{x} + \frac{z}{x} = 3$. By $AM$-$GM$ inequality,$\frac{x^2}{yz} + \frac{y}{x} + \frac{z}{x} \ge 3 \sqrt[3]{\frac{x^2}{yz} \cdot \frac{y}{x} \cdot \frac{z}{x}} = 3(1) = 3$. Equality holds if $\frac{x^2}{yz} = \frac{y}{x} = \frac{z}{x}$,which implies $x=y=z$.$III.$ $x^3+y^2z+z^2x=3xyz$. Let $x=1, y=2, z=0.5$. Then $1 + (4)(0.5) + (0.25)(1) = 1 + 2 + 0.25 = 3.25 
eq 3(1)(2)(0.5) = 3$. However,testing values like $x=1, y=1, z=1$ gives $1+1+1=3$. This condition does not force $x=y=z$ for all positive reals.$IV.$ By $AM$-$GM$,$\frac{x+y+z}{3} \ge \sqrt[3]{xyz}$. Thus $(x+y+z)^3 \ge 27xyz$. Equality holds if and only if $x=y=z$.Thus,$I, II,$ and $IV$ imply $x=y=z$.

Let $x, y, z$ be positive real numbers. Which of the following conditions imply $x=y=z$?
$I.$ $x^3+y^3+z^3=3xyz$
$II.$ $x^3+y^2z+yz^2=3xyz$
$III.$ $x^3+y^2z+z^2x=3xyz$
$IV.$ $(x+y+z)^3=27xyz$

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Let $x, y, z$ be positive real numbers. Which of the following conditions imply $x=y=z$?$I.$ $x^3+y^3+z^3=3xyz$$II.$ $x^3+y^2z+yz^2=3xyz$$III.$ $x^3+y^2z+z^2x=3xyz$$IV.$ $(x+y+z)^3=27xyz$