If (x + y + z) = 0, then prove that (x3 + y3 + z3) = 3xyz.
x+y+z = 0
⇒x+y = -z
(x+y)3= (-z)3
x3+y3+3xy(x+y)= -z3
x3+y3+3xy(-z)= -z3 [x+y= -z]
x3+y3-3xyz= -z3
x3+y3+z3= 3xyz
Hence, x+y+z= 0
⇒x3+y3+z3= 3xyz
If (x + y + z) = 0, then prove that (x3 + y3 + z3) = 3xyz.
x+y+z = 0
⇒x+y = -z
(x+y)3= (-z)3
x3+y3+3xy(x+y)= -z3
x3+y3+3xy(-z)= -z3 [x+y= -z]
x3+y3-3xyz= -z3
x3+y3+z3= 3xyz
Hence, x+y+z= 0
⇒x3+y3+z3= 3xyz