<s>[INST] If $x^9-x^5+x-2=0$ , how to prove $\sqrt[11]{3}If $x^9-x^5+x-2=0$ is known, 
how to prove   $\sqrt[11]{3}<x<\sqrt[10]{3}$ ，$\sqrt[7]{2}<x<\sqrt[6]{2}$。
 [/INST] Hint: $x^9-x^5+x=2 \implies \dfrac{((x^4)^3+1)}{x^4+1}=\dfrac{2}{x}$
Spoiler:
Wolframalpha gives only real solution to this, i.e $x \approx 1.112$, which is in between $3^{\frac{1}{11}}$ and $3^{\frac{1}{10}}$ OR(as Thomas Andrews says) $2^{\frac{1}{6}}$ and $3^{\frac{1}{7}}$  .
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