9. To what power must the root $x_{0}$ of the equation $x^{11} + x^{7} + x^{3} = 1$ be raised to obtain the number $x_{0}^{4} + x_{0}^{3} - 1 ?$
Answer: 15.

Solution. If $x_{0}=1$, then $x_{0}^{4}+x_{0}^{3}-1=1$, so in this case the degree can be any. But the number $x_{0}=1$ does not satisfy the equation $x^{11}+x^{7}+x^{3}=1$, therefore $x_{0} \neq 1$.

Since $1=x_{0}^{11}+x_{0}^{7}+x_{0}^{3}$, we get

$$
x_{0}^{4}+x_{0}^{3}-1=x_{0}^{4}+x_{0}^{3}-x_{0}^{11}-x_{0}^{7}-x_{0}^{3}=x_{0}^{4}\left(1-x_{0}^{7}-x_{0}^{3}\right)=x_{0}^{4}\left(x_{0}^{11}+x_{0}^{7}+x_{0}^{3}-x_{0}^{7}-x_{0}^{3}\right)=x_{0}^{15} .
$$