If x+y+z=0
then prove that
x cube +y cube +z cube=3xyz
Answers
Answered by
105
Hi ,
We know the identity ,
x^3 + y^3 + z^3
= ( x + y + z ) ( x^2 + y^2 + z^2 -xy -yz - xz ) + 3xyz --( 1 )
According to the given problem ,
x + y + z = 0 put this value in equation ( 1 )
x^3 + y^3 + z^3
= 0 ×( x^2 + y^2 + z^2 - xy - yz - zx ) + 3xyz
x^3 + y^3 + z^3 = 3xyz
I hope this helps you.
***
We know the identity ,
x^3 + y^3 + z^3
= ( x + y + z ) ( x^2 + y^2 + z^2 -xy -yz - xz ) + 3xyz --( 1 )
According to the given problem ,
x + y + z = 0 put this value in equation ( 1 )
x^3 + y^3 + z^3
= 0 ×( x^2 + y^2 + z^2 - xy - yz - zx ) + 3xyz
x^3 + y^3 + z^3 = 3xyz
I hope this helps you.
***
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