If $\alpha$ is a real root of the equation $x^5-x^3+x-2=0$, find the value of $\lfloor\alpha^6\rfloor$, where $\lfloor x\rfloor$ is the least positive integer not exceeding $x$.  My approach is to bound the value of $\alpha^6=\alpha^4-\alpha^2+2\alpha$. First I proved the function $f(x)=x^5-x^3+x-2$ is monotone increasing by derivative. Then I argue that $1&lt;\alpha&lt;\frac32$ as $f(1)&lt;0$ and $f(\frac32)&gt;0$. Then I tried to create an upper and lower bound for $\alpha$, as such $$\alpha^6=\alpha^4-\alpha^2+2\alpha&lt;\frac94\alpha^2-\alpha^2+2\alpha=\frac54\alpha^2+2\alpha&lt;\frac{45}{16}+3=\frac{93}{16}$$ and $$\alpha^6=\alpha^4-\alpha^2+2\alpha&gt;\alpha^3-\alpha^2+2\alpha=\alpha^2(\alpha-1)+2\alpha&gt;2$$ Now we know that $\lfloor\alpha\rfloor\in\{2,3,4,5\}$. But I cannot proceed any further. Any idea, help, hint, or answer is appreciated. Thank you.
If $\alpha$ is a real root of the equation   $$x^5-x^3+x-2=0$$   find the value of $\lfloor {\alpha}^6 \rfloor$.      This one totally stumped me. We are asked to calculate $\lfloor {\alpha}^6 \rfloor$ without actually calculating the root or using wolfram alpha or any other calculator. I found that the above equation has only one real root by sketching its graph using calculus. I was also able to use the intermediate value theorem to conclude that $1&lt;\alpha&lt;2$ , but this is of little use while calculating $\lfloor {\alpha}^6 \rfloor$. Please help!
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