Root in Box

Let $f(x) = x^5-x^3+x-2$ , We note that $f'(x) = 5x^4-3x^2+1 > 0$ for all real $x$ . Therefore, $f(x)$ is an increasing function for all $x$ and $f(x)=0$ has only one real solution $\alpha$ . Since $f(1) = -1$ and $f(2) = 24$ , $\implies 1 < \alpha < 2$ .

Since $\alpha$ is a root, then

$\begin{aligned} \alpha^5-\alpha^3+\alpha-2 & = 0 \\ \alpha^5-\alpha^3+\alpha & = 2 \\ \alpha^4-\alpha^2+1 & = \frac 2\alpha \\ (\alpha^2+1)(\alpha^4-\alpha^2+1) & = \frac 2\alpha(\alpha^2+1) \\ \alpha^6+1 & = 2\alpha + \frac 2\alpha \\ \implies \alpha^6 & = 2\alpha + \frac 2\alpha - 1 \end{aligned}$

Now, let $g(\alpha) = 2\alpha + \dfrac 2\alpha -1$ . Then $g'(\alpha) = 2 - \dfrac 2{\alpha^2}$ . We note that $g'(\alpha) > 0$ for $\alpha > 1$ . This means that $g(\alpha)$ is an increasing function for $\alpha >1$ . Since $1 < \alpha < 2$ ,

$\begin{aligned} 2(1) + \frac 21 - 1 < \alpha^6 & < 2(2) + \frac 22 - 1 \\ 3 < \alpha^6 & < 4 \\ \implies \lfloor \alpha^6 \rfloor & = \boxed{3} \end{aligned}$