If x + y + z = 0 , prove that x 2 </msup>

Frank Day

Frank Day

Answered question

2022-07-04

If x + y + z = 0, prove that x 2 2 x 2 + y z + y 2 2 y 2 + z x + z 2 2 z 2 + x y = 1
A problem in my homework had asked me:
When x + y + z = 0, evaluate
x 2 2 x 2 + y z + y 2 2 y 2 + z x + z 2 2 z 2 + x y
Without too much difficulty, one can see that the value should be 1 using ( x , y , z ) = ( 1 , 0 , 1 )
I decided to use x = y z, which turned out not to be as difficult as initially thought. However, would someone care to enlighten me to some other methods of doing this?

Answer & Explanation

billyfcash5n

billyfcash5n

Beginner2022-07-05Added 17 answers

HINT:
2 x 2 + y z = 2 ( y + z ) 2 + y z = ( 2 y + z ) ( y + 2 z )
Now 2 y + z = x + y + z + y x = y x
ban1ka1u

ban1ka1u

Beginner2022-07-06Added 5 answers

The first term is
x 2 2 x 2 x y y 2 = x 2 ( 2 x + y ) ( x y ) = x 2 ( x z ) ( x y )
Do the same transformation on each of the fractions and add them up.
So the whole expression is
x 2 ( x z ) ( x y ) + y 2 ( y z ) ( y x ) + z 2 ( z y ) ( z x ) = . . . = 1

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