(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.