If x + Y + Z =0 show that x cube + y cube + Z cube = 3 x y z

1.	If x + Y + Z =0 show that x cube + y cube + Z cube = 3 x y z
Answer» x^3+y^3+z^3=3xyzx^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. (because(x+y+z)=0)Therefore, x^3+y^3+z^3=3xyz. Hence proved ? TO PROVE : x^3 + y^3 + z^3 =3xyzGIVEN : x+y+z =0PROOF : [ USING 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)] {GIVEN : x+y+z=0} x^3 + y^3 + z^3 - 3xyz = 0 {Any number multiplied by 0 the result is 0} x^3 + y^3 + z^3 = 3xyz {Any number is in negative on L.H.S it result as positive on R.H.S} Hence , proved that x^3 + y^3 + z^3 =3xyz { NOTIFICATION : The mark ^ represent raise to power.}

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