suppose $x\ge y$ $3x+3y+4=x^3+z^3\ge y^3+z^3=3z+3x+4$ $\to y \ge z$ Similar inferred: $x \ge y \ge z \ge x \to x=y=z$ $\to 2x^3=6x+4$<=>$x=-1$ or $x=2$