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Question

Let x, y, z be positive reals. Which of the following implies x=y=z ?
(I) x3+y3+z3=3xyz
(II) x3+y2z+yz2=3xyz
(II) x3+y2z+z2x=3xyz
(II) (x+y+z)3=27xyz

A
I, IV only
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B
I, II, IV only
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C
I, II and III only
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D
All of them
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Solution

The correct option is B I, II, IV only
(I) x3+y3+z3=3xyz
x3+y3+z33xyz=0
(x+y+z)(x2+y2+z2xyyzzx)=0
12(x+y+z)(2x2+2y2+2z22xy2yz2zx)=0
12(x+y+z)[(xy)2+(yz)2+(zx)2]=0
Since x+y+z0,
(xy)2+(yz)2+(zx)2=0
x=y=z

Alternate: AM GM
x3+y3+z333x3y3z3
x3+y3+z33xyz
But, if x3+y3+z3=3xyz
x=y=z [AM=GM]

(II) x3+y2z+yz233x3y2zyz2
x3+y2z+yz23xyz
Then, x3+y2z+yz2=3xyz x=y=z

(III) Suppose x3+y2z+z2x=3xyz (1)x3+y2z2+y2z2+z2x44x3y2zy2zz2x4
3xyz4xyz2 [From (1)]
Therefore, xyz

(IV) x+y+z33xyz
(x+y+z)327xyz
Then, (x+y+z)3=27xyz x=y=z

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