Question

if x^2+y^2+z^2-xy-yz-zx=0, prove that x=y=z

Answer 1

Hey Juhi !

Here is your solution :

Given,

=> x^2 + y^2 + z^2 - xy - yx -zx =0

By multiplying both sides by 2 ,

=> 2 ( x^2 + y^2 + z^2 - xy - yx - zx ) = 0×2

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

Rearranging the terms :

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

Using identity :

[ ( a^2 + b^2 - 2ab ) = ( a - b )^2 ]

=> ( x - y )^2 + ( x - z )^2 + ( y - z )^2 = 0


Let assume that , ( x - y ) ; ( y - z ) and ( x - z ) are non-zero ( nz ) integers.

We also know that square of any integer is always positive( +ve ).

Now,

=> ( x - y )^2 + ( x - z )^2 + ( y - z )^2 = 0

=> ( nz )^2 + ( nz )^2 + ( nz )^2 = 0

=> +ve + ve + ve = 0

It is not possible that the sum of any positive number is 0.

Hence, our assumption that ( x - y ), ( x - z ) , ( y - z ) is non-zero number is wrong.

Therefore, ( x- y ) , ( x - z ) and ( y - z ) are 0.



Now,

=> ( x - y ) = 0

=> ( x - y ) = 0

=> x = y ------ ( 1 )

Again ,

=> ( y - z ) = 0

=> ( y - z ) = 0

=> y = z ------- ( 2 )

From ( 1 ) and ( 2 ), we get ;

=> x = y = z

Proved.

Hope it helps !!

Answer 2

Given,

=> x^2 + y^2 + z^2 - xy - yx -zx =0

By multiplying both sides by 2 ,

=> 2 ( x^2 + y^2 + z^2 - xy - yx - zx ) = 0×2


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

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

=> ( x - y )^2 + ( x - z )^2 + ( y - z )^2 = 0




Now,

=> ( x - y )^2 = 0

=> ( x - y ) = 0

=> x = y ------ ( 1 )

Again ,

=> ( y - z )^2 = 0

=> ( y - z ) = 0

=> y = z ------- ( 2 )

From ( 1 ) and ( 2 ), we get

=> x = y = z

Proved.


Hope it helps !