If X3+y3+z3 =3xyz prove x+y+z=0
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Answered by
1
its so simple
x3+y3+z3=3xyz and x+y+z=0 its an identity
x3+y3+z3=3xyz and x+y+z=0 its an identity
Answered by
6
Given,
x³+y³+z³=3xyz
x³+y³-3xyz=-z³
x³+y³-3xy(x+y)= -z³
(x+y)³=(-z)³
x+y=-z
x+y+z=0
(or)
x+y+z=0
x+y=-z
cubing on both sides
(x+y)³=(-z)³
x³+y³+3xy(x+y)=-z³
x³+y³+z³=3xyz
x³+y³+z³=3xyz
x³+y³-3xyz=-z³
x³+y³-3xy(x+y)= -z³
(x+y)³=(-z)³
x+y=-z
x+y+z=0
(or)
x+y+z=0
x+y=-z
cubing on both sides
(x+y)³=(-z)³
x³+y³+3xy(x+y)=-z³
x³+y³+z³=3xyz
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