9,10,11 |

Solve the equation $x^{3}+x-2=0$ by trial and error and using Cardano's formula.
The equation has an obvious root $x=1$. Dividing by $x-1$, we obtain the quadratic equation $x^{2}+x+2=0$, which has no real roots.

Cardano's formula gives $x=\sqrt[3]{-1+\sqrt{\frac{28}{27}}}+\sqrt[3]{-1-\sqrt{\frac{28}{27}}}$.

From this, it follows that $\sqrt[3]{-1+\sqrt{\frac{28}{27}}}+\sqrt[3]{-1-\sqrt{\frac{28}{27}}}=1$.

## Answer

$x=1$.