Question

Prove that : x^3+y^3+z^3-3*x*y*z =1/2(x+y+z) (x-y)^2+(y-z)^2+(z-x)^2

Answer 1

Heya! here is ya answer

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LHS

x³ + y³ + z³ - 3xyz

= { ( x + y )³ - 3xy ( x + y ) + z³ - 3xyz

= ( x + y + z ) { ( x + y )² - ( x + y )z + z² } - 3xy ( x + y ) - 3xyz

= ( x + y + z ) ( x² + y² + 2xy - xz - yz - z² ) - 3xy ( x + y ) - 3xyz

= ( x + y + z ) { ( x² + y² + z² + 2xy - xz - yz ) - 3xy

= ( x + y + z ) { ( x² + y² + z² - xy - yz - zx ) }

= 1/2 ( x + y + z ) { ( x² + y² - 2xy ) + ( y² + z² - 2yz ) + ( z² + x² - 2zx )

= 1/2 ( x + y + z ) { ( x - y )² + ( y - z )² + ( z - x )² }

RHS

thanks

if you found any mistakes then please tell me.

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