Question

If x+y+z=0, show that x³+y³+z³=3xyz ​

Answer 1

Given

x+y+z=0. ------(1)

then

x+y= -z

cubing on both side

(x+y)^3= -z^3

formula =(x+y)^3= x^3+3xy(x+y)+ y^3

so here also

x^3+3xy(x+y)+y^3 = -z^3

x^3+3xy( -z)+y^3= -z^3

hence

x^3+ y^3+ z^3== 3xyz

Hence proved.~♥~

Answer 2

Answer:

since , X + y + z = 0

➡️ X + y = -z

➡️ ( X + y )³ = (-z)³

➡️ x³+y³+3xy(X+y) = -z³

➡️ x³+y³+3xy(-z) = -z³ ( since X + y = -z)

➡️ x³+y³-3xyz= -z³

➡️ x³+y³+z³=3xyz

hence if X + y + z = 0 then x³+y³+z³= 3xyz