- #1

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I tried a lot, but cant get any answer. Can someone please help me out?

- Thread starter suhasm
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- #1

- 11

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I tried a lot, but cant get any answer. Can someone please help me out?

- #2

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Try:

[tex]x^2 + (y + z)x + (y^2 + z^2 - yz) = 0[/tex] and use the quadratic formula and see where that leads.

--Elucidus

- #3

arildno

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[tex]\frac{1}{2}x^{2}+\frac{1}{2}x^{2}+\frac{1}{2}y^{2}+\frac{1}{2}y^{2}+\frac{1}{2}z^{2}+\frac{1}{2}z^{2}-xy-xz-zy=0[/tex]

and see if you can make some clever manipulations.

- #4

Mark44

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--Elucidus

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y = x + a, and z = x+b.

and ended up with

a^2 + b^2 = ab, where you have to prove that a = b = 0.

- #7

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A much easier way,

y = x + a, and z = x+b.

and ended up with

a^2 + b^2 = ab, where you have to prove that a = b = 0.

from

x2+y2+z2=xy+yz+zx

you can get

2x2+2y2+2z2=2xy+2yz+2zx

which equals

2x2+2y2+2z2-2xy-2yz-2zx=0

,the equivalant is

(x-y)2+(y-z)2+(z-x)2=0.

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