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Let $x, y, z$ be positive reals. Which of the following implies $x=y=z$ ?
$I.$ $x^3+y^3+z^3=3 x y z$
$II.$ $x^3+y^2 z+y z^2=3 x y z$
$III.$ $x^3+y^2 z+z^2 x=3 x y z$
$IV.$ $(x+y+z)^3=27 x y z$
$I, IV$ only
$I, II$ and $IV$ only
$I, II$ and $III$ only
All of them
Solution
(b)
We have, $x=y=z, x, y, z$ positive reals.
$I.$ $x^3+y^3+z^3=3 x y z$
We know,
$x^3+y^3+z^3-3 x y z=(x+y+z)$
$\left(x^2+y^2+z^2-x y-y z-z x\right)$
$= \frac{1}{2}(x+y+z)\left[(x-y)^2\right.\left.+(y-z)^2+(z-x)^2\right]$
When $x=y=z$
Then, $(x-y)^2+(y-z)^2+(z-x)^2=0$
$\therefore \quad x^3+y^3+z^3=3 x y z$
$II$. $x^3+y^2 z+y z^2=3 x y z$
Put $x=y=z$
Then, $LHS = RHS$
$III.$ Put $x=z=1$ and $y=2$
Then, it is also true.
So, we cannot say only for $x=y=z$ for true
$IV. (x+y+z)^3=27 x y z$
$x=y=z$
Then, $(3 x)^3=27 x^3$
Hence option $(iv)$ is also true.
