Question

verify that x cube + y cube + z cube - 3 xyz =1/2(x+y+z) [(x-y) square +(y-z) square +( z-x) square]​

Answer 1

Answer:

r.h.s=1/2(X+y+z)[(x-y)²+(y-z)²+(z-x)²]

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

=X+y+z/2(2x²+2y²+2z²-2xy-2yz-2zx)

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

=2x³+2xy²+2xz²-2x²y-2xyz-2zx²/2+2x²y+2y³+2yz²-2xy²-2y²z-2xyz/2+2x²z+2y²z+2z³-2xyz-2yz²-2z²x/2

=2(x³+xy²+xz²-x²y-xyz-zx²)/2+2(x²y+y³+yz²-xy²-y²z-xyz)/2+2(x²z+y²z+z³-xyz-yz²-z²x)/2

=x³+xy²-xy²+xz²-xz²-x²y+x²y-xyz+xyz-zx²+x²z+y³+yz²-yz²-y²z+y²z+z³

=x³+y³+z³

=l.h.s proved