Question

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

Answer 1

we know that,

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

put x+y+z=0

therefore, 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

hence proved

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Answer 2

Given, x3 + y3 + z3 = 3xyz

Therefore, x3 + y3 + z3 - 3xyz =0

This means,

x3 + y3 + z3 - 3xyz = (x + y + z) (x2 + y2 + z2 - xy - yz - zx)

Now if x + y + z = 0, then......

x3 + y3 + z3 - 3xyz = ( 0 ) (x2 + y2 + z2 - xy - yz - zx)  . . . . . . .... ..... ... .. . . .. . .....    [ subsituting the value of x + y + z ]

x3 + y3 + z3 - 3xyz = 0   

x3 + y3 + z3 = 3xyz .