Polynomial Again?

Actually there are two solutions! And both of them are correct, it means that they hold all of given conditions. Here's my solution:

Those equations can be written as $\begin{aligned} x^3 & = 2+xyz \\ y^3 & = 6+xyz \\ z^3 & = 20+xyz \end{aligned}$ Let $A=xyz$ . Multiply all of equations written above, we have $A^3=(2+A)(6+A)(20+A),$ which is finally a quadratic equation, whose solutions are $A=-4,-\frac{15}{7}$ . Instead of multiplying, if we sum them up, we have $x^3+y^3+z^3=28+3A$ , which can be either $16$ or $\frac{151}{7}$ . Therefore, the answers are $17$ and $158$ .

To ensure that both solutions satisfy all condition, if we have $A=-4$ , then $x^3=2+A=-2$ , and we get $x=-\sqrt[3]{2}$ . Similarly, we get $y=\sqrt[3]{2}$ , and $z=\sqrt[3]{16}$ . We see that $xyz=-4$ , so, everything works fine. Moreover, if $A=-\frac{15}{7}$ , we have $x=-\sqrt[3]{\frac{1}{7}}$ , $y=\sqrt[3]{\frac{27}{7}}$ , $z=\sqrt[3]{\frac{125}{7}}$ . As a result, $xyz=-\frac{15}{7}$ . This also holds.

Edit: if we want the greatest $x^3+y^3+z^3$ , the answer is $158$ .

There are in fact $\color{#D61F06}{\text{two}}$ solutions to the problem.

$\begin{cases} x^3 - xyz = 2 & \Rightarrow x^3 - 2 = xyz \\ y^3 - xyz = 6 & \Rightarrow y^3 - 6 = xyz \\ z^3 - xyz = 20 & \Rightarrow z^3 - 20 = xyz \end{cases}$ .

$\begin{aligned} \Rightarrow x^3 - 2 & = y^3 - 6 = z^3 - 20 \\ \Rightarrow y^3 & = x^3 + 4 \\ z^3 & = x^3 + 18 \end{aligned}$

Therefore,

$\begin{aligned} x^3 - xyz & = 2 \\ x^3 - x\sqrt[3]{(x^3+4)(x^3+18)} &= 2 \\ x^3 - 2 & = x \sqrt [3] {(x^3+4)(x^3+18)} \\ (x^3 - 2)^3 & = x^3 (x^3+4)(x^3+18) \\ x^9 - 6x^6 + 12x^3 - 8 & = x^9 + 22x^6 + 72x^3 \\ 28x^6 + 60x^3 + 8 & = 0 \\ 7x^6 + 15x^3 + 2 & = 0 \\ (7x^3 + 1)(x^3+2) & = 0 \end{aligned}$

$\Rightarrow \begin{cases} x^3 = -\frac{1}{7} & \Rightarrow x^3 + y^3 +z^3 = -\frac{1}{7} + -\frac{1}{7} + 4 + -\frac{1}{7} + 18 = \frac{151}{7} & \Rightarrow m + n = \boxed{158} \\ x^3 = -2 & \Rightarrow x^3 + y^3 +z^3 = -2 - 2 + 4 - 2 + 18 = 16 & \Rightarrow m + n = 17 \end{cases}$