Question

B) If x+y+z=0, Show that xcube +ycube +z cube =3xyz​

Answer 1

Answer:

dy9dt9xoy yonytxyctxudiyfoytdidt8fiyify8yfljcxkhvjllxhoxu6

Step-by-step explanation:

td7fyifiyfgifhtudvho5est7ts7w47e7t

Answer 2

Answer:

x³+y³+z³=3xyz proved

Step-by-step explanation:

We know that,

x³+y³+z³-3xyz= (x+y+z)(x²+y²+z²-xy-yz-zx)

putting x+y+z=0,

x³+ y³+z³-3xyz=(0)(x²+y²+z²-xy-yz-zx)

x³+y³+z³-3xyz=0

Now,

x³+y³+z³=3xyz

hence proved