10.6. Determine all integers $n$ for which the number $A=\sqrt[3]{n+\sqrt[3]{n-1}}$ is rational.
Solution: Let $A \in \square$. Then $\exists x \in \square: x^{3}=n+\sqrt[3]{n-1} \cdot \sqrt[3]{n-1}=x^{3}-n \in \square \Rightarrow \exists y \in \square: y^{3}=n-1$. $n=y^{3}+1$. Therefore $x^{3}=y^{3}+y+1$.

If $n>1$, then $y>0$. Hence $y^{3}<x^{3}=y^{3}+y+1<y^{3}+3 y^{2}+3 y+1=(y+1)^{3}$. Since $x, y \in \square$ the inequality $y^{3}<x^{3}<(y+1)^{3}$ is impossible.

If $n=1$, then $A=1 \in \square$.

If $n=0$, then $A=-1 \in \square$.

If $n<0$, then $y<-1$. Hence $(y-1)^{3}=y^{3}-3 y^{2}+3 y-1<y^{3}+y+1=x^{3}<y^{3}$. Since $x, y \in \square$ the inequality $(y-1)^{3}<x^{3}<y^{3}$ is impossible.

Answer: $n=0, n=1$.