Question

if x+y+z=0 show that x3+y3z3​

Answer 1

Answer:

3xyz

Step-by-step explanation:

x³+y³+z³

=(x+y)³-3xy(x+y)+z³

= (x+y)³+z³ - 3x²y - 3xy²

= (x+y+z)³-3(x+y)z(x+y+z)- 3x²y - 3xy²

= (x+y+z)³-3(xz+yz)(x+y+z)- 3x²y - 3xy²-3xyz+3xyz

= (x+y+z)³-3(xz+yz)(x+y+z)- 3xy(x+y+z)+3xyz

= (x+y+z){(x+y+z)²-3xy-3yz-3xz}+3xyz

= (x+y+z)(x²+y²+z²-xy-yz-xz)+3xyz

if x+y+z=0;  (x+y+z)(x²+y²+z²-xy-yz-xz) = 0

so, x³+y³+z³= 3xyz