if X + Y + Z = 0 show that x cube + y cube + Z cube equals to 3
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Answered by
1
To prove we can take an example
X=2, Y=3,Z=4
X^3+Y^3+Z^3=3xyz
3*2*3*4=72
There fore by x^3+y^3+z^3=3xyz
rahulthegenius84:
nice method to do
thank you!
My pleasure
Answered by
2
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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