# Prove that x=y=z if x^2+y^2+z^2-xy-yz-zx=0

2
by Nitya8

• Brainly User
2014-10-31T23:00:54+05:30
Here's what you got, x²+y²+z²-xy-yz-zx=0. Multiply this equation with 2. You get this, 2x²+2y²+2z²-2xy-2yz-2zx=0. Rearrange the terms in this manner, (x²-2xy+y²) + (y²-2yz+z²) + (z²-2zx+x²)=0. You have, (x-y)²+(y-z)²+(z-x)²=0. Here's a little brain work you need to do. You might be knowing that the square of a real number is either zero or positive. Now, see that the terms on the L.H.S. are all squares of real numbers ( if you think of x, y, z to be real numbers instead of complex, which you forgot to mention ) and the R.H.S. is 0. Hence, each term has to be 0, if this were not so the L.H.S. would be positive. So you have, x-y=0 and y-z=0 and z-x=0 giving you x=y=z.
If x, y, z, aren't real, this result wouldn't apply.
2014-11-01T10:09:39+05:30

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X² + y² + z² -xy - xz - yz = 0
multiplying 2 on both sides

2x² + 2y² + 2z² - 2xy - 2xz - 2yz = 0
x² + y² - 2xy+ x² + z² - 2xz + z² + y² - 2yz  = 0
(x - y)² + (x - z)² + (y - z)² = 0

so the  value are in square so they can't be negative so they must be indivudually zero to satisfy the equation

so x - y = 0 ⇒ x = y
x - z = 0 ⇒ x = z
y - z = 0 ⇒ y = z

so hence x = y = z proved