Question

verify that x cube + y cube + Z cube minus 3 x y z is equal to one half (X + Y + z )(x-y)^2 + (Y - z)^2 + (z-x )^2

Answer 1

Heya, friend

Your Solution Given in the above attachment ....

×××××× Verified××××××
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Answer 2

Hey mate here is your answer.........

This answer is verified ✅✅✅

We have :
$( {x}^{3} + {y}^{3} + {z}^{3} - 3xyz) = (x + y + z)( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - zx)$
$= ( ({x + y}^{3} ) - 3xy(x + y)) + {z}^{3} - 3xyz$
$= {u}^{3} - 3xyu + {z}^{3} - 3xyz \: \: where \: (x + y) = u$
$= ( {u}^{3} + {z}^{3} ) - 3xy(u + z)$
$= (u + z) - 3xy(u + z)$
$= (u + z)( {u}^{2} - uz + {z}^{2} ) - 3xy(u + z)$
$= (u + z)( {u}^{2} + {z}^{2} - uz - 3xy)$
$= (x + y + z)(( { x + y}^{2} ) + {z}^{2} - (x + y)z - 3xy)$
$= (x + y + z)( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - zx)$
$proved.....$
hope it will help you.....