Question

If x^2+y^2+z^2-xy-yz-zx=0 then prove x=y=z.

Answer 1

Here, the given equation,

x^2 + y^2 + z^2 - xy - yz - zx=0

\implies \frac{1}{2}(2x^2 + 2y^2 + 2z^2 - 2xy - 2yz - 2zx)=0

\frac{1}{2}(x^2 - 2xy + y^2 + y^2 + z^2 - 2yz + z^2 - 2zx + x^2)=0

\frac{1}{2}[(x-y)^2 + (y-z)^2 + (z-x)^2]=0

\implies (x-y)^2 + (y-z)^2 + (z - x)^2 = 0----(1)

∵ If A² + B² + C²= 0

⇒ A = B = C = 0

( because square of a number can not be negative )

Thus, from equation (1),

x-y = y - z = z - x = 0

⇒ x = y = z

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