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(1) \(xyz=-6\). Infinitely many combinations of \(x\), \(y\) and \(z\) are possible which will give different values of the expression in the stem: try \(x=y=1\) and \(y=-6\) or \(x=1\), \(y=2\), \(z=-3\). Not sufficient.

(2) \(x+y+z=0\). Rearrange: \(x=-(y+z)\) and substitute this value of \(x\) into the expression in the stem to get \(\frac{x^3+y^3+z^3}{xyz}=\frac{-(y+z)^3+y^3+z^3}{xyz}=\frac{-y^3-3y^2z-3yz^2-z^3+y^3+z^3}{xyz}=\frac{-3y^2z-3yz^2}{xyz}=\frac{-3yz(y+z)}{xyz}\), as \(x=-(y+z)\) then: