Question

If x2+y2+z2-xy-yz-zx=0
Then,prove that x=y=z

Answer 1

x²+y²+z² - xy - yz - zx = 0

muntply the equation by 2/2

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

we know that when we do square of a number then we get only positive numbers .so here we can see that the square of three numbers is = 0
and normally it is not possible....but when the numbers are individually= 0
so we the numbers= 0

x-y = 0
x = y

y-z =0
y= z

z-x = 0
z= x

so x = y = z

Answer 2

Answer:

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

Hence, proved....