It's not an "Unpredictable Solution" Problem!

$\begin{cases} 2x+y+xy = 2 \\ y+ 3z + yz = 7 \\ 2x + 3z + xz = 30 \end{cases}$

---

The first equation, $\begin{aligned} 2x + y + xy & = 2 \\ y + xy & = 2-2x \\ y(x+1) & = 2- 2x \\ y & = \dfrac{2-2x}{x+1} \end{aligned}$

---

The third equation, $\begin{aligned} 2x + 3z + xz & = 30 \\ xz + 3z & = 30-2x \\ z(x+3) & = 30-2x \\ z & = \dfrac{30-2x}{x+3} \end{aligned}$

---

The second equation, $\begin{aligned} y + 3z + yz & = 7 \\ y + 3z + yz + 3 & = 10 \\ (y+3)(z+1) & = 10 \\ \left( \color{#D61F06} \dfrac{2-2x}{x+1} \color{#333333} + 3 \right)\left( \color{#3D99F6} \dfrac{30-2x}{x+3} \color{#333333} + 1 \right) & = 10 \\ \left( \color{#D61F06} \dfrac{2-2x+3x+3}{x+1} \color{#333333} \right)\left( \color{#3D99F6} \dfrac{30-2x + x+3}{x+3} \color{#333333} \right) & = 10 \\ \left( \dfrac{ \color{#E81990} x+5}{\color{#20A900} x+1} \color{#333333} \right)\left( \dfrac{\color{#E81990} 33-x}{\color{#20A900} x+3} \color{#333333} \right) & = 10 \\ \color{#E81990} 28x - x^{2} + 165 & \color{#333333} = 10( \color{#20A900} x^{2} + 4x + 3) \\ 28x - x^{2} + 165 & = 10x^{2} + 40x +30 \\ 11x^{2} + 12x -135 & = 0 \\ (11x+45)(x-3) & = 0 \end{aligned}$

---

After all, we have $x= 3$ ; $\Rightarrow \color{#EC7300} \text{ Since } \space x \space \text{ is an integer}$

Substituting, we have $y = -1$ and $z= 4$ .

Now, we have $\sqrt[4]{x^{3} + 10y^{3} + z^{3}} = \sqrt[4]{3^{3} + 10(-1)^{3} + 4^{3}} = \sqrt[4]{81} = \pm 3$ .

The positive value of $\sqrt[4]{x^{3} + 10y^{3} + z^{3}}$ is $\boxed{3}$ .

Relevant wiki: Quadratic Diophantine Equations - Solve by Simon's Favorite Factoring Trick

$\begin{cases} 2x+y+xy = 2 & \implies (x+1)(y+2) = 4 & ...(1) \\ y+3z+yz = 7 &\implies (y+3)(z+1) = 10 & ...(2) \\ 2x+3z+xz = 30 & \implies (x+3)(z+2) = 36 & ...(3) \end{cases}$

From $(1): \implies y + 2 = \dfrac 4{x+1}$ . From $(3): \implies z+2 = \dfrac {36}{x+3}$ . Then, from $(2)$ :

$\begin{aligned} (y+3)(z+1) & = 10 \\ \left(\frac 4{x+1} + 1\right) \left(\frac {36}{x+3}-1\right) & = 10 \\ (4+x+1)(36-x-3) & = 10(x+1)(x+3) \\ (x+5)(33-x) & = 10(x^2+4x+3) \\ -x^2 + 28x + 165 & = 10x^2 + 40x + 30 \\ 11x^2 + 12x - 135 & = 0 \\ (11x+45)(x-3) & = 0 \\ \implies x & = 3 & \small \color{#3D99F6} x \text{ is an integer.} \end{aligned}$

When $x=3$ , $(1): y + 2 = \dfrac 4{x+1} \implies y = -1$ , $(3): z+2 = \dfrac {36}{x+3} \implies z = 4$ .

$\implies \sqrt[4]{x^3+10y^3+z^3} = \sqrt[4]{27-10+64} = \sqrt[4]{81} = \boxed{3}$