Question

verify that x³+y³+z³-3xyz=½(x+y+z)[(x-y)²+(y-z)²+(z-x)²]​

Answer 1

Answer:

R.H.S.= ½(x+y+z)[(x-y)²+(y-z)²+(z-x)²]

= ½(x+y+z)[ x²-2xy+ y²+y²-2yz+ z²+z²-2zx+ x²]

= ½(x+y+z)*2*( x²+y²+z²-2xy-2yz-2zx )

= (x+y+z)( x²+y²+z²-2xy-2yz-2zx )

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

=x³+xy²+z²x-2x²y-2xyz-2zx²+y³+x²y+yz²-2xy²-2xyz-2zy²+z³+y²z+zx²-2z²y-2xyz-2z²x

2xyz-2zy²+z³+y²z+zx²-2z²y-2xyz-2z²x

= x³+y³+z³-3xyz .....= L.H.S.

therefore,

L.H.S.= R.H.S

hope it works out for you...