Question

x+y+z=0 show that x cube+ycube+zcube+3xyz=?​

Answer 1

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Answer 2

Answer:

If x+y+z=0, then x^3+y^3+z^3=3xyz

Step-by-step explanation:

Algebraic Identity: x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)

Substituting x+y+z=0 in the above identity.

x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)

x^3+y^3+z^3-3xyz= 0 (x^2+y^2+z^2-xy-yz-zx)

x^3+y^3+z^3-3xyz=0

x^3+y^3+z^3=3xyz (shifting -3xyz to right hand side)

Hence proved!