Question

if (x+y+z)=0,show thatx3+y3+z3=3xyz​

Answer 1

x³+y³+z³+3xy+3yz+3xz

Answer 2

x³+y³+z³ = 3xyz

Then ,

x³+y³+z³-3xyz = 0

Expanding the identity :

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

And then it is given that x+y+z = 0

Putting this value in the above identity ,

(0)(x²+y²+z²-xy-yz-xz)

= 0

Hence it is proved that the above identity is true !!

x³+y³+z³ = 3xyz (If x+y+z = 0)