(x+y+z)² + (x – y+z)^2+(x + y- z)^2
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Given equations are x² + y² + z² = 525
xy + yz + zx = 50
Per algebraic identity.
(x + y + z )² = x² + y² + z² + 2xy + 2yz + 2zx
(x + y + z )² = x² + y² + z² + 2(xy+yz+zx)
= 525+2(50)
= 625
x+y+z = √625
= ±25
After squareroot will be cancelled , the number become ±25.
( x + y + z )² = x² + y² + z² + 2xy +2yz +2zx
( ±25 )² = 525 + 2( 50 )
625 = 525 + 100
625 = 625
L.H.S = R.H.S.
So , the answer is ' ±25 '.
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