Let x, y, z be positive reals. Which of the f | vedclass

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(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-

Vedclass · Accepted Answer

$I, II$ and $IV$ only

Vedclass · Answer

(b)

We have, $x=y=z, x, y, z$ positive reals.

$I.$ $x^3+y^3+z^3=3 x y z$

We know,

$x^3+y^3+z^3-3 x y z=(x+y+z)$

$\left(x^2+y^2+z^2-x y-y z-z xight)$

$= \frac{1}{2}(x+y+z)\left[(x-y)^2ight.\left.+(y-z)^2+(z-x)^2ight]$

When $x=y=z$

Then, $(x-y)^2+(y-z)^2+(z-x)^2=0$

$	herefore \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.

Let $x, y, z$ be positive reals. Which of the following implies $x=y=z$ ?

$I.$ $x^3+y^3+z^3=3 x y z$

$II.$ $x^3+y^2 z+y z^2=3 x y z$

$III.$ $x^3+y^2 z+z^2 x=3 x y z$

$IV.$ $(x+y+z)^3=27 x y z$

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