If x = 997, y = 998, z = 999, then the value of x2 + y 2

If x = 997, y = 998, z = 999, then the value of x² + y ² + z ² – xy – yz – zx will be

Here , x = 997, y = 998, z = 999
Given in the question ,
x² + y ² + z ² – xy – yz – zx
On Multiplying and dividing by 2 , we get

=	2	( x² + y ² + z ² – xy – yz – zx )
	2

=	1	( x² + y ² + z ² – 2xy – 2yz – 2zx )
	2

=	1	( x² + y² - 2xy + y² + z² - 2yz + x² + z² - 2zx )
	2

=	1	[ ( x -y )² + ( y - z )² + ( z - x )² ]
	2

=	1	[ ( 997 - 998 )² + ( 998 - 999 )² + ( 999 - 997 )² ]
	2

=	1	[ 1² + 1² + 2² ]
	2

=	1	× 6 = 3
	2

x² + y ² + z ² – xy – yz – zx = 3