Question

If x³+y³+z³ = 3xyz, then which one of the following is true ?

(1) x+y+z =1

(2) x-y+z = 0

(3) either x+y+z = 0 or x=y=z

(4) neither x+y+z = 0 or x=y=z

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{know}\mathrm{that}, \\ {x}^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right) \\ \mathrm{When}x+y+z=0,\mathrm{then}\mathrm{the}\mathrm{above}\mathrm{identity}\mathrm{becomes}, \\ {x}^3+y^3+z^3-3xyz=0\left(x^2+y^2+z^2-xy-yz-zx\right) \\ \Rightarrow x^3+y^3+z^3-3xyz=0 \\ \Rightarrow x^3+y^3+z^3=3xyz \end{gathered}$$