Question

x 3 +y 3 +z 3 -3xyz = 1/2 (x+y+z)[(x-y) 2 +(y-z) 2 +(z-x) 2 ]

Answer 1

rhs = 1/2(x+y+z)[(x-y)²+(y-z)²+(z-x)²]
=1/2(x+y+z)[x²+y²-2xy+y²+z²-2yz+z²+x²-2zx]
=1/2(x+y+z)[2x²+2y²+2z²-2xy-2yz-2zx]
=1/2(x+y+z)*2*[x²+y²+z²-xy-yz-zx]
=(x+y+z)(x²+y²+z²-xy-yz-zx)
=x³+y³+z³-3xyz
=lhs

Answer 2

Answer:

here u go ur answer

Step-by-step explanation:

1/2(x+y+z)[(x-y)²+(y-z)²+(z-x)²]

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

=1/2(x+y+z)[2x²+2y²+2z²-2xy-2yz-2zx]

=1/2(x+y+z)*2*[x²+y²+z²-xy-yz-zx]

=(x+y+z)(x²+y²+z²-xy-yz-zx)

=x³+y³+z³-3xyz

hence proved